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... , bi_ 1, xi, bi+ 1, ... , bM) 1/Ji(xj) = gi(b 1, ... , bi_ 1, xi, bi+ 1, ... , bM),
i
= 1, 2, ... , N.
7.1. Partial Derivatives and the Chain Rule
177
Then, according to Equation (7.9), we have
tp(xi) = f[l/l 1 (xi), I/J 2 (xj), ... , 1/JN(xi)]. With h denoting a real number, define
litp = tp(bj + h) - tp(bj) t;..pi = .pi(bj +h)- 1/Ji(bj),
i = 1, 2, ... , N.
Since each .pi is continuous at bi, it follows that t;..pi--+ 0 ash--+ 0. We apply Equation (7.4) of Theorem 7.2 to the function tp and use t;..pi instead of h; in that equation. We obtain (7.10) In this formula, we have 8; = 8;(lii/Jl, ... , fit/IN) and 8;--+ 0 as t;..pk--+ 0, k = 1, 2, ... , N. Now write Equation (7.10) in the form
fitp
h
N
= i~
fii/Ji
{f;[g1(b), ... , gN(b)]
+ 8;}h,
valid for lhl sufficiently small. Letting h tend to 0, we get the statement of the Chain rule. D
Definitions. Let D be a subset of ~Nand suppose f: D--+ ~ 1 is a given function. For x = (x 1 , ... , xN) and k a real number, we use the symbol kx to denote (kx 1 , kx 2 , ••• , kxN). Thenfis homogeneous of degree n if and only if(i) kx e D whenever x e D and k -::!- 0, and (ii) f(kx)
= k"f(x) for x e D, k -::!- 0.
The function f is positively homogeneous of degree n if and only if the two conditions above hold for all k > 0 and all x e D.
Remarks. The quantity n need not be an integer. For example, the function f: (x, y)--+ x- 113 + y- 1' 3 is homogeneous of degree -1/3. A function may be positively homogeneous but not homogeneous. The function f: (x, y) --+ x 2 + y 2 is positively homogeneous of degree 1 but not homogeneous.
J
Theorem 7.4 (Euler's theorem on homogeneous functions). Suppose that f: ~N--+ ~ 1 is positively homogeneous of degree nand suppose that !, 1 , f. 2 , ••• , fN are continuous for a-::!- 0. Then N
L aJ;(a) =
i=1
The proof is left to the reader.
nf(a).
(7.11)
7. Differentiation in IRN
178
PROBLEMS
1. Let D be an open connected set in IRN and suppose f: D-+ IR 1 has the property that f 1 = f 2 = · · · = fN = 0 for all xED. Show that f =constant in D.
2. Let f: IRN -+ IR 1 be given and suppose that g 1, g 2 , ••• , gN are N functions from IRM into IR 1 • Let h 1 , h2 , ••• , hM be M functions from Winto IR 1 . Give a formula for the Chain rule for H,;(x), where
g 2 [h 1 (x), ... , hM(x)], ... }.
H(x) = f{g 1 [h 1 (x), ... , hM (x)], 3. Write a proof of Theorem 7.2 for N = 3.
4. Use the Chain rule to compute H, 2 (x) where His given by (7.9) and f(x) =xi + x~- 3x 3 , g 1 (x) =sin 2x 1 + x 2 x 3 , g 2 (x) =tan x 2 + 3x 3 , g 3 (x) = x 1 x 2 x 3 •
5. Use the Chain rule to compute H, 3 (x) where H is given by (7.9) and f(x) = 2x 1 x 2 + x~- x~, g 1 (x) = log(x 1 + x 2 ) - x~, g 2 (x) =xi+ x~, g 3 (x) = xix 3 + x 4 , g 4 (x) = cos(x 1 + x 3 ) - 2x 4 . 6. Consider the function f: IR 2 f( X
1, X 2
-+
IR 1 defined by
)= {
: 1x 2 X1
+ X 22
,
0,
(x 1 ,x 2 )i=(O,O), X1
= X 2 = 0.
(a) Show that f is not continuous at x 1 = x 2 = 0. (b) Show that f 1 and f 2 exist at x 1 = x 2 = 0. Why does this fact not contradict Theorem 7.2? 7. Consider the function f: IR 2 -+ IR 1 such that f 1 and f 2 exist and are bounded in a region about x 1 = x 2 = 0. Show that f is continuous at (0, 0). 8. Given the function f: IR 2
Show that f 1 and f continuous at (0, 0).
2
-+
IR 1 defined by
are bounded near (0, 0) and therefore (Problem 6) that f is
9. Iff: IRN-+ IR 1 is homogeneous of degree 0, show by a direct computation that f satisfies Euler's differential equation: N
I X;f; = i::;:l
0.
10. Prove Theorem 7.4, Euler's theorem on homogeneous functions.
7.2. Taylor's Theorem; Maxima and Minima Definitions. Let f be a function from !RN into IR 1 • We define the second partial derivative J.;.i as the first partial derivative of f.; with respect to xi. We define
7.2. Taylor's Theorem; Maxima and Minima
179
the third partial derivative !,;,j,k as the first partial derivative of f.;.i· Fourth, fifth, and higher derivatives are defined similarly. In computing second partial derivatives it is natural to ask whether the order of computation affects the result. That is, is it always true that f.;.i = f.i.i for i =f. j? There are simple examples which show that the order of computation may lead to different results. (See Problem 3 at the end of this section.) The next theorem gives a sufficient condition which validates the interchange of order of partial differentiation.
Theorem 7.5. Let f: !RN ~ IR 1 be given and suppose that f, f.;, f.i.i• and f.i.i are all continuous at a point a. Then
f · .(a) = f · ·(a) • 1,)
PROOF. We establish the result for N the general case is exactly the same. Writing a= (a 1 , a 2 ), we define
Ad= f(a 1 +
,),1
•
= 2 with i = 1 and j = 2. The proof in
h, a 2 +h)- f(a 1 + h, a 2)- f(a 1, a 2 +h)+ f(a 1, a 2).
(7.12)
We shall show that Ad/h 2 tends to the limit f. 1 • 2 (a 1 , a 2 ) and also that the same quantity tends to f. 2 , 1 . Hence the two second derivatives must be equal. Define (7.13) cp(s) = f(a 1 + s, a 2 + h) - f(a 1 + s, a 2), 1/l(t)
= f(a 1 +
h, a 2 + t)- f(a 1, a 2 + t).
(7.14)
Then
Ad= cp(h) -
cp(O),
Ad =
1/l(h) - 1/1(0).
(7.15)
We apply the Mean-value theorem to Equations (7.15), getting
Ad= 1/l'(tdh,
(7.16)
where s 1 and t 1 are numbers between 0 and h. From Equations (7.13) and (7.14), it follows that
cp'(sd = /. 1(a 1 + s 1, a2 +h)- f. 1 (a 1 + s 1, a 2),
(7.17)
1/l'(td = /. 2(a 1 + h, a 2 + t 1)- f. 2(a 1, a 2 + td.
(7.18)
A second application of the Mean-value theorem to Equations (7.17) and (7.18) yields cp'(sl) = !.1,2(a1 + s1, a 2 + t2)h, 1/l'(tl) = f.2.1(a1 + s2, a2 + tdh, where s 2 and t 2 are numbers between 0 and h. Substituting these expressions in Equations (7.16) we find
1 h 2 Ad= f.1.2(a1 + s1, a2 + t2) = f.2.1(a1 + S2, a2 + td.
7. Differentiation in IRN
180
Letting h tend to zero and observing that s 1 , s 2 , t 1 , and t 2 all tend to zero with D
h, we obtain the desired result.
Definitions. A multi-index rx is an element (rx 1 , rx 2, ... , rxN) in IRN where each rx; is a nonnegative integer. The order of a multi-index, denoted by lrxl, is the nonnegative integer rx 1 + rx 2 + · · · + rxN. We extend the factorial symbol to multi-indices by defining rx! = rx 1 ! · rx 2 ! ... rxN!. If f3 is another multi-index (/31, f3z, ... , f3N), we define rx + f3 = (rx1 + fJ1, rxz + f3z, ... , rxN + f3N). Let x = (x 1 , x 2 , by the formula
••• ,
xN) be any element of!RN. Then the monomial x" is defined
Clearly, the degree of x" is lrxl. Any polynomial in IRN is a function f of the form (7.19) CaX" f(x) =
L
lal
x 2, ... , xN) is an
nl
L ~x'" l«i=n ex! ~
n!
..
..
(7.20)
£.... I IXl' ... XNN• «, + ... +«N=n (Xl. • • • (XN•
-
PROOF. We fix the integer nand prove Equation (7.20) by induction on the integer N. For N = 1, Equation (7.20) becomes
which is true. Now, suppose that Equation (7.20) holds for N = k. We shall prove that it also holds for N = k + 1. To do this, we first observe that the binomial theorem yields (x 1 + x 2 +
··· + xk+ 1)" = [(x 1 + ··· + xk) + xk+ 1]" ~
= £.... •1( i=O 1·
n!
.
.) xl+1(X1
n- 1 1.
.
+ ··· + xkr-1.
(7.21)
From the induction hypothesis, the right side of Equation (7.21) becomes II
~
!--
1=0
Setting cxk+l becomes
=j
~
n.I
(
')I n _ 1.
..
.. i
£.... . 'I( -")I I lxl' ... xkkxk+l· «,+···+«k=n-J1· n 1 · CX1····cxk.
(7.22)
and cancelling (n- j)!, we find that Expression (7.22)
This last expression is the right side of Equation (7.20) with k replaced by k + 1. The induction is established. D Let G be an open set in IRN and letf: G-+ IR 1 be a function with continuous second derivatives in G. We know that in the computation of second derivatives the order of differentiation may be interchanged. That is, J.i,J = !, 1,i for all i and j. We may also write f.i.J = DlDd] and, using the symbol o for composite maps as described in Theorem 6.40, we have for two differential maps Di and D1 Di o D1(f) = D;[D1f].
Iff has third and higher order derivatives, then D; o D1 o Dk(f) = Di{D1[Dk(f)]}, D; o D1 o Dt o D1(f) = D;(D1{Dt[D1(f)]} ), and so on. We shall usually omit the symbol
o,
especially when the order of
7. Differentiation in IRN
182
differentiation may be interchanged. A linear combination of differential maps is called a differential operator. If a, b, c, d are constants, the combination aD1 D2 D3
+ bD2 D 1 D 1 + cD1 D3 + dD2
is an example of a differential operator. A differential operator acts on functions which are assumed to have continuous derivatives up to the required order on a fixed open set in IRN. With any polynomial in IRN of the form P(el,e2•····eN)=
L c,.e . ,
(7.23)
L
(7.24)
l11l.;n
we associate the differential operator P(D 1 , D2 ,
.•• ,
DN) =
c,.D11•
l11l.;n
If oc is the multi-index (oc 1, oc 2, .•• , ocN), then D11 is the operator given by D 11 = D~· Di 2 ••• Dil· That is, D 11f means that f is first differentiated with respect to xN exactly ocN times; then it is differentiated ocN-l times with respect to xN-1> and so on until all differentiations of f are completed. The order of the differential operator (7.24) is the degree of the polynomial (7.23). By induction it is easy to see that every differential operator consisting of a linear combination of differential maps is of the form (7.24). The differentiations may be performed in any order. The polynomial P(el> e2, ... ,eN) in (7.23) is called the auxiliary polynomial of the operator (7.24). We illustrate the use of the notation with an example. 2. Suppose that f: IR 2 -.!R 1 is given by f(x 1 , x 2 ) = x 1 e"'' cos x 2 • Let P(e 1 , e 2) = et + 2eie~ + et be a given polynomial. Show that P(D 1 , D2 )f = 0. EXAMPLE
Solution. We have
D1 (f) =
aaf
= (x 1
+ l)e"'• cos x 2 ;
xl
= (x 1
v:(f) = (xl
+ 2)e"'' cos x 2 ; + 3)e"'' cos x2;
D2 (f) = -x 1 e"'• sin x 2 ;
Di(f) = -x 1 e"'' cos x 2 ;
Di(f) = x 1 e"'' cos x 2 •
Therefore,
D We now introduce several definitions and simple facts concerning the algebra of linear operators with the purpose of applying them to linear differential operators as defined by (7.24). These operators are useful in the
7.2. Taylor's Theorem; Maxima and Minima
183
proof of Taylor's theorem and the second derivative test for maxima and minima of functions of several variables (Theorem 7.8 below).
Definitions. Let L 1 , L 2 , ••• , Lt be differential operators, each of which has the same domain and range in a Euclidean space. Let c 1 , c 2 , ... , ck be real numbers. The operator denoted by c 1 L 1 + c2 L 2 + · · · + ckLk is the operator L such that L(f) = c 1L 1 (f) + ·· · + ctLt(f) for all fin the domain of the L 1, i = 1, 2, ... , k. The operator L 1 L 2 is the operator L such that L(f) = L 1 {L 2(f)} for all fin the domain of L 2 and with the function L 2 (f) in the domain of L 1 • The operators L 1 L 2L 3 , L 1 L 2L 3 L 4 , etc., are defined similarly. Lemma 7.3. Suppose that L 1 , L 2 , ••• , Lk are differential operators and c 1 , c 2 , ••. , Ct are real numbers. Let P;(~) be the auxiliary polynomial for Li> i = 1, 2, ... , k. Then
(i) the auxiliary polynomial c 1 Pt(~)
P(~)
for c 1L 1
+ c2 L 2 + ··· + CtLt
is P(~) =
+ ··· + ckPk(~);
(ii) the auxiliary polynomial for L 1 L 2 ••• Lt is P1 (~)P2 (~) ... Pt(~). PROOF. Let n be the maximum order of all the operators L 1• Then we may write
Li=
L
1«1 ....
i=1,2, ... ,k.
bi 0. If f"(a) = 0, the test fails. With the aid of Taylor's theorem for functions from ~N into ~ 1 we can establish the corresponding result for functions of N variables. Definitions. Let f: ~N-+ ~ 1 be given. The function f has a local maximum at the value a if and only if there is a ball B(a, r) such that f(x) - f(a) ~ 0 for x E B(a, r). The function f has a strict local maximum at a if and only if f(x) - f(a) < 0 for x E B(a, r) except for x = a. The corresponding definitions for local minimum and strict local minimum reverse the inequality sign. Iff has partial derivatives at a, we say that f has a critical point at the value a if and only if DJ(a) = 0, i = 1, 2, ... , N. Theorem 7.8 (Second derivative test). Suppose that f: ~N-+ ~ 1 and its partial derivatives up to and including order 2 are continuous in a ball B(a, r). Suppose that f has a critical point at a. For h = (h 1 , h 2 , .•• , hN), define llf(a, h)= f(a + h) - f(a); also define Q(h) =
1
1
N
L ----,D«.f(a)h« = 12. 1,]=1 -~ DiDif(a)hihi. 1«1=2
(7.31)
(X.
(a) If Q(h) > 0 for h =F 0, then 1 f has a strict local minimum at a. (b) If Q(h) < 0 for h =F 0, then f has a strict local maximum at a. (c) If Q(h) has a positive maximum and a negative minimum, then llf(a, h) changes sign in any ball B(a, p) such that p < r.
PRooF. We establish Part (a), the proofs of Parts (b) and (c) being similar. Taylor's theorem with remainder (Theorem 7.7) for n = 1 and x =a+ his f(a +h)= f(a) +
L
1«1=1
D«J(a)h~~. +
L
1
1 D«J(e)h~~.,
1«1=2 oc.
(7.32)
e
where is on the straight line segment connecting a with a + h. Since f has a critical point at a, the first sum in Equation (7.32) is zero, and the
1 If Q is a quadratic form such as Q(h) = L~J=l aiihihJ, then Q is positive definite if and only if Q(h) > 0 for all h "# 0. Also, Q is negative definite when - Q is positive definite.
187
7.2. Taylor's Theorem; Maxima and Minima
second sum may be written L1f(a, h)= Setting lhl 2 = h~
L
AD"l"(a)h'" +
1«1=2 0!.
L
A[D'1(e)- D'1(a)]h'".
1«1=2 0!.
(7.33)
+ h~ + ··· + h~ and e(h) =
L
1«1=2
A[D'1(e>- D'1(a)J lhh'"12 ,
(X.
we find from Equation (7.33) L1f(a, h) = Q(h) + lhl 2 e(h).
(7.34)
Because the second partial derivatives off are continuous near a, it follows that e(h) -+ 0 as h -+ 0. Also, 2 ~ hi hj 2 Q(h) = lhl t,f;t DPJf(a)fhj"fiiT = lhl Qt(h).
The expression Q 1 (h) is continuous for h on the unit sphere in RN. According to the hypothesis in Part (a), Q 1 , a continuous function, must have a positive minimum on the unit sphere which is a closed set. Denote this minimum by m. Hence, Q(h) ~ lhl 2 m for all h. Now choose Ihi so small that le(h)l < m/2. Inserting the inequalities for Q(h) and le(h)l into Equation (7.34), we find L1f(a, h)> tlhl 2 m for lhl sufficiently small and h ¥- 0. We conclude that minimum at a.
f
has a strict local D
Remarks. The quantity Q given in Expression (7.31) is a quadratic form. In linear algebra, we develop the fact that Q is positive definite if and only if the matrix (AiJ) (DPJf(a)) has all positive eigenvalues. Also, Q is negative definite when all the eigenvalues of (A 1J) are negative. The quadratic form Q has a positive maximum and a negative minimum when the matrix has at least one positive and at least one negative eigenvalue. It is not necessary to find all the eigenvalues in order to determine the properties of Q. It is sufficient to "complete the square", as in elementary algebra, to determine which of the cases (a), (b), or (c) of Theorem 7.8 prevails.
=
PROBLEMS
1. Given f: IR 3 -+ IR 1 defined by f(x1, x2, x 3) =(xi+ x~ + x~)- 112, Show that f satisfies the equation /.1,1
+ /.2,2 + /.3,3 =
0.
7. Differentiation in IRN
188
2. Given f: IR 2 -+ IR 1 defined by f(x 1 , x 2) = xt- 2x~x 2 - x 1 x 2 and given L 1 (D) = 2D1 - 3D2 , L 2(D) = D1 D2, show that (L 1 L 2 )(/) = (L2Ld(f). 3. Given f: IR 2
Show that /
-+
IR 1 defined by
1 , 2 (0,
0)
= -1 and /
2 , 1 (0,
0)
= 1.
4. Write out the proof of Theorem 7.5 for a function f from IRN into IR 1 . 5. Prove the Binomial theorem (Lemma 7.1). 6. Establish the Taylor expansion for functions from IR 1 into IR 1 (Theorem 7.6). [Hint: Make use of the function f'(t)(x - t) J(a) q>(x) = 0.]
=
7. Write out explicitly all the terms of the Taylor expansion for a function/: IR 3 -+ IR 1 for the case n = 2. 8. Find the relative maxima and minima of the function f: IR 2
f(x 1 , x2) = x~
IR 1 given by
+ 3x 1 x~- 3xf- 3x~ + 4.
9. Find the critical points of the function f: IR 4 f(x 1 , x 2 , x 3 , x 4 )
-+
-+
IR 1 given by
= xf + x~ + x~- xi- 2x 1 x 2 + 4x 1 x 3 + 3x 1 x 4 - 2x 2x 4
+ 4x 1 - 5x 2 + 7.
10. Write out proofs of Parts (b) and (c) of the Second derivative test (Theorem 7.8).
In each of Problems 11 through 13, determine whether Q: R 3 definite, negative definite, or neither. 11. Q(x 1 , x 2 , x 3 )
= xf + 5x~ + 3x~- 4x 1 x 2 + 2x 1 x 3 -
12. Q(xl> x 2 , x 3 ) = xf 13. Q(x 1 , x 2 , x 3 )
+ 3x~ + x~- 4x 1 x 2 + 2x 1 x 3
= -xf- 2x~- 4x~- 2x 1 x 2 -
-
-+
IR 1 is positive
2x 2 x 3 6x 2 x 3
2x 1 x 3
7.3. The Derivative in IRN Each partial derivative of a function f: RN-+ IR 1 is a mapping from RN into R 1• This generalization of the ordinary derivative, useful for many purposes, is unsatisfactory in that it singles out a particular direction in which the differentiation is performed. We now take up another extension of the ordinary derivative, one in which the difference quotient tends to a limit as x-+ a regardless of the direction of approach.
7.3. The Derivative in IRN
189
Let A be an open subset of ~N, and suppose that f and g are functions from A into ~ 1 • We denote by d(x, y) the Euclidean distance between two points X,
yin
~N.
Definition. The continuous functions f and g are tangent at a point a E A if and only if lim lf(x) - g(x)l = O. d(x, a) x-+a xi' a
We note that if two functions are tangent at a point a then, necessarily, f(a) = g(a). Also, iff, g, and hare functions with f and g tangent at a and with g and h tangent at a, then we verify simply that f and h are tangent at a. To see this, observe that lf(x) - h(x)l d(x, a)
----::-c-----:-- ~
lf(x) - g(x)l d(x, a)
+
lg(x) - h(x)l . d(x, a)
As x --+ a, the right side tends to 0 and hence so must the left side. Let L: ~N--+ ~ 1 be a linear function; that is, Lis of the form N
L(x) = b0
+L
k=1
bkxk,
where b0 , b1 , •.. , bN are real numbers. It may happen that a function f is tangent to a linear function at a point a. If so, this linear function is unique, as we now show.
Theorem 7.9. Suppose that L 1 and L 2 are linear functions tangent to a function fat a point a. Then L 1 = L 2 • PROOF.
It is convenient to write the linear functions L 1 and L 2 in the form N
N
L 1 (x) = c 0
+L
k=1
ck(xk- ak),
L 2 (x)
=
c~
+L
From the discussion above, it follows that L 1 (a) c0 = c~. Also, for every e > 0, we have
k=1
cl.(xk - ak).
= L 2 (a) = f(a).
Hence, (7.35)
for all x sufficiently close to a. For z E ~N, we use the notation liz II Now, with () a real number, choose (jz
x-a=w·
Then, for sufficiently smalllbl, we find from Inequality (7.35) that liz I = elbl. Ikf:1 (ck- ck)fzlf I~ elbl W N
, (jzk
= d(z, 0).
7. Differentiation in IRN
190
Therefore,
and since this inequality holds for all positive k = 1, 2, ... , N.
6,
= clc,
we must have ck
D
Definitions. Suppose that f: A-+ IR 1 is given, where A is an open set in IRN containing the point a. The function f is differentiable at a if and only if there is a linear function L(x) = f(a) + L:=l ck(xk - ad which is tangent to fat a. A function f is differentiable on a set A in IRN if and only if it is differentiable at each point of A. The function L is called the tangent linear function to f at the point a. The function L is also called the derivative or total derivative off at a. We use the symbol f'(a) for the derivative off at the point a in IRN. Theorem 7.10. Iff is differentiable at a point a, then f is continuous at a. The proof is left to the reader. As we saw in Section 7.1, a function/may have partial derivatives without being continuous. An example of such a function is given in Problem 6 at the end of Section 7.1. Theorem 7.10 suggests that differentiability is a stronger condition than the existence of partial derivatives. The next theorem verifies this point.
Theorem 7.11. Iff is differentiable at a point a, then all first partial derivatives off exist at a.
PROOF. Let L be the tangent linear function to fat a. We write N
L(x) = f(a)
+
L ck(xk -
k=l
ak).
From the definition of derivative, it follows that /(x)- f(a)-
lim
1
I
ck(xk- ak)l
= 0.
k=l
d(x, a)
x-+a x;fa
For x we choose the element a Equation (7.36) becomes lim if(a h;-+0
h;;fO
+ h where h = + h) - f(a) lhd
(7.36)
(0, 0, ... , 0, hi, 0, ... , 0). Then
cihd = O;
7.3. The Derivative in IRN
191
we may therefore write l
f(a +h)- f(a) _ hi
·I=
C,
8
(h·) I'
(7.37)
where e(h1) -+ 0 as h1 -+ 0. We recognize the left side of Equation (7.37) as the expression used in the definition of f,;(a). In fact, we conclude that C·l
= f·(a) ,l '
i = 1, 2, ... , N.
0
A partial converse of Theorem 7.11 is also true.
Theorem 7.12. Suppose that f.;, i = 1, 2, ... , N are continuous at a point a. Then f is differentiable at a. This result is most easily established by means of the Taylor expansion with remainder (Theorem 7. 7) with n = 0. We leave the details for the reader.
Definitions. Suppose that f has all first partial derivatives at a point a in IRN. The gradient off is the element in IRN whose components are (f.l (a), f.2(a), ... , f.N(a)).
We denote the gradient off by Vf or grad f Suppose that A is a subset of IRN and f: A-+ IR 1 is differentiable on A. Let h = (h 1 , h2 , ••• , hN) be an element of IRN. We define the total differential df as that function from A x IRN -+ IR 1, given by the formula N
df(x, h) =
L f.k(x)hk.
(7.38)
k=l
Using the inner or dot product notation for elements in IRN, we may also write
df(x, h) = Vf(x) ·h. Remarks. Equation (7.38) shows that the total differential (also called the differential) of a function f is linear in h and bears a close resemblance to the tangent linear function. The differential vanishes when x = a and h = 0, while the tangent linear function has the value f(a) at the corresponding point. The Chain rule (Theorem 7.3) takes a natural form for differentiable functions, as the following theorem shows.
Theorem 7.13 (Chain rule). Suppose that each of the functions g 1, g2 , ••• , gN is a mapping from IRM into IR 1 and that each g 1 is differentiable at a point b = (b1 , b2 , •.• , bM). Suppose that f: IRN-+ IR 1 is differentiable at the point a= (g 1 (b), g2 (b), ... , gN(b)). Form the function H(x) = f[g 1 (x), g2 (x), ... , gN(x)].
7. Differentiation in [RN
192
Then H is differentiable at b and N
dH(b, h) =
L j;[g(b)] dgi(b, h).
i:l
The proof is similar to the proof of Theorem 7.3, and we leave it to the reader.
PROBLEMS
1. Let f and g be functions from IRN into IR 1 • Show that iff and g are differentiable at a point a, then f + g is differentiable at a. Also, rx.f is differentiable at a for every real number rx.. 2. Let f: IR 3
->
IR 1 be given by
Find Vf and df(x, h).
3. Let f and g be functions from IRN into IR 1 • Show that iff and g are differentiable, and rx. and p are numbers, then V(rx.f + pg) = rx.Vf + pVg, V(fg) = fVg
+ gVf
-> IR 1 be given. Assume that the range off is contained in the domain of g. Show that, in such a case,
4. Let f: IRN-> IR 1 and g: IR 1
V[g(f)]
= g'Vf
5. Show that iff is differentiable at a point a, then it is continuous at a. (Theorem 7.10.) 6. Suppose that all first partial derivatives of a function f: IRN .... IR 1 are continuous at a point a. Show that f is differentiable at a. (Theorem 7.12.) 7. Given f: IRN-> IR 1 and an element be IRN with lib II = 1. The directional derivative off in the direction b at the point a, denoted by d&f, is the function from [RN into IR 1 defined by . f(a + tb) - f(a) db f =hm .
t-o
t
Suppose that all first partial derivatives off are continuous at a point a. Show that the directional derivative off in every direction exists at a and that
(d&f)(a) = Vf(a) ·b. 8. Suppose that A is an open set in IRN. Let f: IRN .... IR 1 be differentiable at each point of A. Assume that the derivative off, a mapping from A into IR 1, is also differentiable at each point of A. Then show that all second partial derivatives off exist at each point of A and that for all i, j = 1, 2, ... , N. 9. Prove Theorem 7.13.
7.3. The Derivative in IRN
193
10. Suppose f and g are functions from IRN into IRM. We denote by dN and dM Euclidean distance in IRN and IRM, respectively. We say that f and g are tangent at a point a E IRN if and only if lim dM(f(x), g(x)) = O. dN(x, a)
x-+a
Show that iff is tangent to a linear function L: IRN --. IRM at a point a, then there can be only one such. [Hint: Write L as a system of linear equations and follow the proof of Theorem 7.9.] 11. (a) Let f: IRN--. IRM be given. Using the result of Problem 10, define the derivative
off at a point a.
(b) Let the components off be denoted by fl, f 2, ... ,JM. Iff is differentiable at a, show that all partial derivatives of Jl, jl, ... , JM exist at the point a. 12. Let L: IRN --.IRM be a linear function. Show that L c6 + Lf=tcixk, j = 1, 2, ... , M.
= (L 1 , L 2 , ••• , LM) where Li =
13. Let A be a closed region in IRN. Suppose thatf: IRN--. IR 1 is differentiable in a region containing A and that /has a maximum value at a point a E oA. Show that d.f ~ 0 at the point a where n is the inward pointing unit normal to oA at the point a.
CHAPTER
8
Integration in ~N
8.1. Volume in IRN In order to construct the theory of Riemann integration in IRN, we need to develop first a theory of volume for sets of points in IRN. This development is a straightforward generalization of the theory of area (Jordan content) given in Section 5.4. We shall outline the main steps of the the theory and leave the proofs of the theorems to the reader. We first recall the definitions of open and closed cells given earlier in Section 6.3.
Definitions. Let a= (a 1 , a2 , ... , aN) and b = (b1 , b2 , ... , bN) be points in IRN with a; < b; for i = 1, 2, ... , N. The set R = 1{ x : a; < X; < b;, i = 1, 2, ... , N} is called an open cell in IRN. A closed cell is the set {x : a; ~ X; ~ b;, i = 1, 2, ... , N}. If the relations
b1
-
a 1 = b2
-
a2 = .. · = b" - a"
hold, then the cell is called a hypercube. Note that for N = 2 an open cell is the interior of a rectangle, and for N = 3 an open cell is the interior of a rectangular parallelepiped. For N = 2 and 3, hypercubes are squares and cubes, respectively. Paralleling the theory given for N = 2, we divide all of IRN into closed hypercubes given by the inequalities
k; k;- 1 --::>:::x.::>:::• ~ 2"' ~ 2"
i = 1, 2, ... , N,
(8.1)
where the k;, i = 1, 2, ... , N, are integers. As we shall see below, the volume of each such hypercube is 1/2"N. 194
8.1. Volume in IRN
195
LetS be a bounded set in IRN. We wish to define the volume of S. Suppose that Sis contained in an open cell R = {x : A; < X; < B;, i = 1, 2, ... , N} where, for convenience, A 1 , ..• , AN, B1 , ... , BN are integers. R denotes the closed cell having the same "sides" as R. Definitions. Suppose that IRN is divided into hypercubes as given by Expression (8.1). Such a system of hypercubes divides IRN into a grid. Each such hypercube entirely contained in the point set S is called an inner cube for S of the nth grid. Hypercubes which contain at least one point of S but are not inner cubes are called boundary cubes. Hypercubes which contain no points of S are called exterior cubes. A hypercube which has each edge of length 1 contains (2n)N = 2Nn hypercubes of the nth grid. Therefore the closed cell R containing S has 2Nn(B1
-
A 1 }(Bz- Az} ... (BN- AN)
hypercubes of the nth grid. We define the quantities
v,.- (S) = v,.+(S)
=
2-Nn times the number of inner cubes for S, v,.-(S) + 2-Nn times the number of boundary cubes for S.
The following lemma is the direct analog of Lemma 5.3. Lemma 8.1. Let R = {x: A;< X;< B;, i = 1, 2, ... , N} be an open cell where A;, B;, i = 1, 2, ... , N, are integers. Suppose that S c R. Then, for every subdivision ofiRN into hypercubes given by (8.1), all inner cubes and all boundary cubes of S are contained in R. Theorem 8.1. Suppose that RandS are as in Lemma 8.1. Then (i) o:::;; v,.-(s):::;; v,.+(s):::;; (B1 - A 1 )(B2 - A 2 ) ••• (BN- AN); (ii) v,.- (S) :::;; V,.~ 1 (S) for each n; (iii) V,.~ 1 (S) :::;; v,.+ (S) for each n; (iv) the sequences {V,.-(S)}, {V,.+(S)} tend to limits as n--. oo. If the limits are denoted by
v- (S) =
lim
v,.- (S)
and
v+ (S) =
lim
v,.+ (S),
then N
0:::;; v-(S):::;; v+(S):::;;
TI (B;- A;); i=l
(v) the quantities
v,.- (S), v,.+ (S), v- (S), v+ (S) are independent of R.
Note that Theorem 5.19 is exactly the same as Theorem 8.1 for the case N = 2. The proof of Theorem 8.1 parallels that of Theorem 5.19 and is left to the reader.
8. Integration in IRN
196
Definitions. The number v- (S) is called the inner volume of S and the number v+ (S) is called the outer volume of S. A set of points S is said to have volume if and only if v- (S) = v+ (S). The volume of S, denoted V(S) (sometimes denoted VN(S) if we wish to emphasize the number of dimensions) is this common value. A set of points in IRN which has a volume is called a figure. Theorem 8.2. Let sl and s2 be bounded sets in RN. Then (i) (ii) (iii)
(iv)
if S1 is contained in S2 , then v- (S1 ) ~ v- (S2) and v+ (Stl ~ v+ (S2); v+ (Stu S2) ~ v+ (St) + v+ (S2); if sl and s2 have no common interior points, then v- (St u S2) ~ v- (St) + v- (S2); if S1 and S2 are figures and if S1 and S2 have no common interior points, then sl u s2 is a figure and V(S1 u S2) = V(S1 ) + V(S2);
(v) the open cell R = {x: a1 < x 1 < b1, i = 1, 2, ... , N} is a figure and N
V(R) =
0
(b1 - a 1).
i=l
Also, R is a figure and V(R) = V(R). The statement of Theorem 8.2 is the same as Theorem 5.20 for N = 2, and the proofs are similar.
Theorem 8.3. Suppose that sl and s2 are figures in RN. Then sl u s2 is a figure. Also, S1 n S2 and S1 - S2 are figures. The boundary of any figure has volume zero. The proof of Theorem 8.3 follows in precise detail the proof of Theorem 5.21 and is left to the reader. Theorems 8.2 and 8.3 are easily extended to the case where sl' S2, ... ' sk is any finite collection of sets in IRN. PROBLEMS
1. Give an example of a bounded set s in IRN for which
v- (S) < v+ (S).
2. Consider the open cell R = {x: a1 < x 1 < b1} in which a 1 , a2 , •• • , aN, b1 , b2 , are irrational numbers. Show that v,.- (R) < v,.+ (R) for every n.
••• ,
bN
3. Prove Lemma 8.1. Is the same result true ifthe {A 1} and {B1} are rational numbers provided that the grid given by the inequalities (8.1) has n sufficiently large? 4. Give an example of a nonempty bounded set S in IRN with the property that v,.- (S) = v,.+ (S) for every n. 5. Prove Theorem 8.1.
197
8.2. The Darboux Integral in IRN 6. State and prove for sets in IRN the analog of Lemma 5.3 of Section 5.4.
7. Give an example oftwo sets sl and s2 in IRN such that neither sl nor s2 is a figure but S1 u S2is a figure and S1 n S2is a figure. Is it possible that S1 - S2is a figure? 8. Prove Theorem 8.2. 9. Prove Theorem 8.3. 10. Show that if Sl, S2, ... , sk are figures in IRN, then sl u s2 u ... u sk and sl n s2 n ... n sk are figures. If sl, ... , sk, ... is an infinite collection of figures, is it true that U::'; 1 s. is a figure?
8.2. The Darboux Integral in
~N
The development of the theory of integration in ~N for N ~ 2 parallels the one-dimensional case given in Section 5.1. For functions defined on an interval I in ~1, we formed upper and lower sums by dividing I into a number of subintervals. In ~N we begin with a figure F, i.e., a bounded region which has volume and, in order to form upper and lower sums, we divide F into a number of subfigures. These subfigures are the generalizations of the subintervals in ~1, and the limits of the upper and lower sums as the number of subfigures tends to infinity yield upper and lower integrals.
Definition. Let F be a bounded set in ~N which is a figure. A subdivision of F is a finite collection of figures, {F1 , F2 , ••• , F11 }, no two of which have common interior points and whose union is F. That is, F
= F1 u
F2 u · · · u F11 ,
Int(F;) n Int(Fj) = 0
for
i =F j.
We denote such a subdivision by the single letter~Let D be a set in ~N containing F and suppose that f: D--+ ~ 1 is bounded on F. Let ~ be a subdivision ofF and set Mi
= sup f
on
F;,
mi
= inf f
on Fi.
Definitions. The upper sum off with respect to the subdivision ~ is defined by the formula II
s+ (f, ~) = I
Mi V(Fi),
j;l
where V(Fi) is the volume in ~N of the figure Fi. Similarly, the lower sum off is defined by II
s_ (f, ~) = L
mi V(F;).
j;l
Let~
be a subdivision of a figure F.
Then~',
another subdivision ofF, is
8. Integration in IRN
198 F
A2 : a subdivision of F
A,: a subdivision ofF
F
A': the common refinement of A1 and A2
Figure 8.1
called a refinement of L\ if and only if every figure of L\' is entirely contained in one of the figures of L\. Suppose that L\ 1 = {F1, F2 , ••• , Fn} and L\ 2 = { G1, G2 , .•• , Gm} are two subdivisions of F. We say that L\', the subdivision consisting of all nonempty figures of the form Fi n Gi, i = 1, 2, ... , n, j = 1, 2, ... , m, is the common refmement of L\ 1 and L\ 2 • Note that L\' is a refinement of both L\ 1 and L\ 2 (see Figure 8.1).
Theorem 8.4. Let F be a figure in
~Nand suppose that f: F-+ ~ 1 is bounded
on F.
if L\ is any subdivision ofF, then mV(F) ~ S_(f, L\) ~ s+(J, L\) ~ MV(F).
(a) If m ~ f(x) ~ M for all x e F and
(b) If A' is a refinement of L\, then
s_ (f, L\) ~ s_ (f, L\')
and
s+ (f, L\') ~ s+ (f, L\).
(c) If L\ 1 and L\ 2 are any two subdivisions ofF, then
s_ (f. L\d ~ s+ (f. L\2). PROOF
(a) The proof of Part (a) is identical to the proof of Part (a) in Theorem 5.1, the same theorem for functions from ~ 1 into ~ 1 • (b) Let L\' = {F~, ... , F,;.} be a refinement of L\ = {F1 , F2 , ••• , Fn}· We denote by F~, F2, ... , F~ the figures of L\' contained in F1 • Then, using the symbols m1 = inf f
in
F1
we have immediately m1
and ~
m~ =
m;, i =
inf f
in F[,
i = 1, 2, ... , k,
1, 2, ... , k, since each F[ is a subset of F1 •
199
8.2. The Darboux Integral in IRN
+ ··· + V(Fk), we have m1 V(Fd ::s:;; mJ.(FD + ··· + m1 V(Fk).
Because V(Fd = V(FD
(8.2)
The same type of inequality as (8.2) holds for F 2 , F3 , ..• , F". Summing these inequalities, we get S_ (f, ~) ::s:;; S_ (f, ~'). The proof that s+ (f, ~') ::s:;; s+ (f, ~) is similar. (c) If ~ 1 and ~ 2 are two subdivisions, let~· denote the common refinement. Then, from Parts (a) and (b), it follows that s_(f, ~1)
::s:;;
s_(f, ~')
::s:;;
s+(f, ~')
::s:;;
Definitions. Let F be a figure in ~Nand suppose that f: F-+ on F. The upper integral off is defined by
L
f dv =
g.I.b. s+ (f.
o
s+(f, ~2). ~1
is bounded
~).
where the greatest lower bound is taken over all subdivisions~ of F. The lower integral off is
L f
dV = l.u.b. S_(f,
~).
where the least upper bound is taken over all possible subdivisions ~ if F. If LfdV= LfdV,
then we say that f is Darboux integrable, or just integrable, on F and we designate the common value by LfdV.
When we wish to emphasize that the integral is N-dimensional, we write LfdVN.
The following elementary results for integrals in ~N are the direct analogs of the corresponding theorems given in Chapter 5 for functions from ~ 1 into ~ 1 • The proofs are the same except for the necessary alterations from intervals in ~ 1 to figures in ~N.
Theorem 8.5. Let F be a figure in
~N.
Let f, f 1 , f 2 be functions from F into ~ 1
which are bounded. (a) If m
::s:;;
f(x)
::s:;;
M for x m V(F)
E
F, then
::s:;;
L f dV
::s:;;
I f dV
::s:;;
MV(F).
8. Integration in IRN
200
kf, then
(b) If k is a positive number and g =
t
t
f dV
and
tgdV= k tfdV
and
g dV = k
t
t -
-
g dV = k
f dV.
If k is negative, then
(c) The following inequalities hold:
L I
(d) If f 1 (x) ~
+ fz) dV ~
f1 dV
+
fz dV;
U1
+ fz) dV
f1 dV
+
f 2 dV.
f 2 (x) for all x
t
f1 (x) dV
L L I ~I
U1
~
E
t
F, then
t -
f 2(x) dV,
-
-
f1(x) dV
~
t -
f 2(x) dV.
(e) Suppose G is another figure in ~N such that F and G have no common interior points. Iff is defined and bounded on F u G, then
I
f dV =
I-
f dV =
-
FuG
I
-
FuG
F
f dV
+
I
-
f dV;
G
I- f dV + I- f dV. F
G
If the functions considered in Theorem 8.5 are Darboux integrable, then Formulas (a)-( e) of that theorem are modified as in the Corollary to Theorem
5.3.
Theorem 8.6. Let F be a figure in
~Nand suppose that
f: F--. ~ 1
(a) f is Darboux integrable on F if and only if for each
B
is bounded.
> 0 there is a
subdivision .::\ of F such that
s+ (f, .::\) - s_ (f, .::\)
to be the set of interior points of F;. It may happen that F/ 0 > is empty for some values of i. For each F/0 >which is not empty, we select a closed figure G; contained in F/0 > such that e i = 1, 2, ... , m. V(F;- G;) < 4Mm' See Figure 8.4 where such a selection is shown with m = 4 for a figure in IR 2 . It is not difficult to verify that such closed figures G1 , G2 , ••• , Gm can always be found. For example, each G; may be chosen as the union of closed hypercubes interior to F; for a sufficiently small grid size. Since each set G; is closed and is contained in F/0 >, there is a positive number {> such that every ball B(x, f>) with x in some G; has the property that B(x, f>) is contained in the corresponding set F/0 >. Let L\ be any subdivision with mesh less than f>. We shall show that the first inequality in (8.6) holds for this subdivision. We separate the figures of L\ into two classes: 11 , 12 , ••• , 1" are those figures of L\ containing points of some G;; K 1 , K 2 , ..• , Kq are the remaining figures of L\. Denote by L\' the common refinement of L\ and L\ 0 • Because of the manner we chose f>, each 1; is contained entirely in some F~ 0 >. Therefore, 1 1 , which in 12 , ••• , 1" are figures in the refinement S. The remaining figures of L\' are composed of the sets K; n F), i = 1, 2, ... , q; j = 1, 2, ... , m. We have the inequality e e m ~ V(F;- G;) < m·-- = - . L V(Kk) < 4M 4Mm i=t k=t
L
We introduce the notation
M; = sup f(x), xeJ1
Mi =
sup f(x), xeK;
M;i
= sup f(x). xeKinFJ
8.3. The Riemann Integral in IRN
207
Using the definitions of s+(f, Ll) and s+(f, Ll'), we find n
s+ (f, .ll) = s+ (f, Ll') =
I
q
I M; V(K;),
+
M; V(J;)
i=l
i=l
n
q
m
I Mij V(K; n FJ). I j=l I M; V(J;) + i=l i=l
Now it is clear that V(K;) = follows that
Lj= 1 V(K; n Fj). Therefore, by subtraction, it m
q
s+(f,.ll)-s+(f,.ll')=
LI
(M;-M;j)V(K;nFj)
i=l j=l
m
q
:::; 2M
L L V(K;nFj) i=l j=l
:::; 2M
= -2. L V(K;) < 2M·~4 M i=l
e
q
e
(8.8)
According to Part (b) of Theorem 8.4, we have
s+ (f, .ll') :::; s+ (f, Llo). Combining this fact with Inequalities (8.7) and (8.8), we conclude that
t
s+ (f, Ll)
0 be given. Then there is a tJ > 0 such that if Ll is any
8. Integration in IRN
208
subdivision ofF x G with mesh less than (), we have s+ (f, Ll)
n,
{!
and suppose that f.. is Riemann integrable for each n. Define the Riemann integral JF f dV = lim...."' JF f.. dV when the limit exists. Lettingf(x, y) = 1/(x 2 + y 2 )" and F = {(x, y): 0 ~ x 2 + y 2 ~ 1} in IR 2 , show that JFf(x, y) dVexists for a< 1 and that it does not for a~ 1.
for unbounded functions by the formula
8. Integration in a;tN
210
5. Let F be a figure in IRN. Show that iff: F-+ IR 1 is Riemann integrable then it is Darboux integrable (Theorem 8.11). 6. Suppose f is defined in the squareS
f (x, Y) -_
= {(x, y): 0 ~ x ~ 1, 0 ~ y ~ 1} by the formula
{1
4y 3
ifxisirrational, · ratmna · 1. 1"f x 1s
(a) Show that JA<JAf(x, y) dy) dx exists and has the value 1. (b) Show that Jsf(x, y) dV does not exist. 7. Suppose that in the square S = {(x, y): 0 ~ x ~ 1, 0 ~ y ~ 1} we define the set A as the set of points (x, y) such that x and y are rational and that, when they are represented in the form of x = p.fq 1 , y = p 2 /q 2 (lowest terms), then q 1 = q 2 • Suppose that f: S-+ IR 1 is given by f( ) = {0 if(x, y) e A, x,y 1 if(x,y)eS-A. (a) Show that JA<JAf(x, y) dy) dx and JA<JAf(x, y) dx) dy, both exist and have the value 1. (b) Show that Jsf(x, y) dV does not exist. 8. Write a proof of the Corollary to Theorem 8.13. 9. Let F be a regular figure in IRN. Suppose f: F-+ IR 1 and g: F-+ IR 1 are Riemann integrable on F. Show thatfg if Riemann integrable on F.
CHAPTER
9
Infinite Sequences and Infinite Series
9.1. Tests for Convergence and Divergence It is customary to use expressions such as u1
+ u2 + ... + Un + ...
00
and
L Un n=l
(9.1)
to represent infinite series. The u; are called the terms of the series, and the quantities
n = 1, 2, ... , are called the partial sums of the series. The symbols in (9.1) not only define an infinite series but also are used as an expression for the sum of the series when it converges. To avoid this ambiguity we define an infinite series in terms of ordered pairs. Definitions. An infinite series is an ordered pair ({un}, {sn}) of infinite sequences in which sn = u 1 + u2 + ··· + un for each n. The un are called the terms of the series and the sn are called the partial sums. If there is a number s such that sn-+ s as n-+ oo, we say the series is convergent and that the sum of the series iss. If the s" do not tend to a limit we say that the series is divergent. It is clear that an infinite series is uniquely determined by the sequence {un} of its terms. There is almost never any confusion in using the symbols in (9.1) for an infinite series. While the definition in terms of ordered pairs is satisfactory from the logical point of view, it does require a cumbersome notation. Rather than have unwieldy proofs which may obscure their essential 211
9. Infinite Sequences and Infinite Series
212
features, we shall use the standard symbols of u 1 + u2 + · · · + un + · · · and un to denote infinite series. The context will always show whether the expression represents the series itself or the sum of the terms. L~= 1
Theorem 9.1. If the series L~= 1 un converges, then un--+ 0 as n--+ oo. PROOF. For all n > 1, we have un = sn- sn_ 1 .1f s denotes the sum of the series, 0 then sn--+ s and sn_ 1 --+ s as n--+ oo. Hence un--+ s - s = 0 as n--+ oo.
Remarks. If the terms un of an infinite series tend to zero, it does not necessarily follow that the series L~= 1 un converges. See the Corollary to Theorem 9.5 which illustrates this point. Let U1 + Uz + ... + Un + ... be a given series. Then a new series may be obtained by deleting a finite number of terms at the beginning. It is clear that the new series will be convergent if and only if the original series is.
Theorem 9.2. Let L~= 1 un, L~= 1 vn be given series and let c -=!= 0 be a constant. (a) If L~= 1 un, L~= 1 Vn are convergent, then L~= 1 (un L~= 1 cun are convergent series. Also, 00
00
00
L Vn, L Un ± n=1 L (un ± Vn) = n=1 n=1
+ vn), L~= 1 (un - vn), and 00
00
L Un. L cun = n=1 n=1 C
(b) If L~= 1 un diverges, then L~= 1 cun diverges.
PROOF. For each positive integer n, we have n n n vk, uk ± (uk ± vk) =
L
k=1
L
L
k=1
k=1
Then Part (a) follows from the theorems on limits. (See Section 2.5.) To prove Part (b), we have only to observe that if L~= 1 cun converges, then so does 0 L~= 1 (1/c)(cun) = L~= 1 un. A series of the form
a + ar
+ ar 2 + ··· + ar" + ···
is called a geometric series. The number r is the common ratio.
Theorem 9.3. A geometric series with a-=!= 0 converges if lrl < 1 and diverges lrl ~ 1. In the convergent case, we have a arn-1 = - - . L 1- r n=1 oo
if
9.1. Tests for Convergence and Divergence
213
PROOF. We verify easily that for each positive integer n
sn = a + ar +
0
0
0
arn a + arn-1 = - - - - - . 1-r 1-r
If lrl < 1, then rn-+ 0 as n-+ oo (see Section 2.5, Problem 6). Hence sn-+ a/(1 - r). If lrl ;;,:: 1, then undoes not tend to zero. According to Theorem D 9.1, the series cannot converge. Theorem 9.4 (Comparison test). Suppose that un ;;,:: 0 for all n. (a) If un ~ an for all n and L~= 1 an converges, then L~= 1 un converges and
L~=1 Un ~ L~=1 an. (b) Let an ;;,:: 0 for all n. If L~= 1 an diverges and un;;,:: an for all n, then L~ 1 un diverges. The proof is left to the reader. (See Problems 11 and 12 at the end of this section.) Let f be continuous on [a, oo). We define
J oo a
f(x) dx = lim b-++oo
fb f(x) dx
(9.2)
a
when the limit on the right exists. The term improper integral is used when the range of integration is infinite. We say the improper integral converges when the limit in (9.2) exists; otherwise the integral diverges. Theorem 9.5 (Integral test). Suppose that f is continuous, nonnegative, and nonincreasing on [1, oo ). Suppose that L~= 1 un is a series with un = f(n), n = 1, 2, ... Then (a) L~= 1 un converges if Sf f(x) dx converges; and (b) L~=1 un diverges if Sf f(x) dx diverges. PROOF. Since f is positive and nonincreasing, we have (see Figure 9.1) for n ;;,:: 2,
I
}=2
u1 ~
u In1 f(x) dx ~ nf j=1
We define F(X) =
1.
Ix
f(x) dx.
If this integral converges, then F(X) is a nondecreasing function which tends to a limit, and so F(n) is a bounded nondecreasing sequence. Thus denoting sn = L}= 2 u1, we see that sn ~ F(n), and sn tends to a limit. Part (a) is now established. If the integral diverges, then F(X)-+ +oo as X-+ oo. Therefore F(n)-+ +oo. Since F(n) ~ LJ:~i u1, we conclude that the series diverges. D
9. Infinite Sequences and Infinite Series
214 y
Figure 9.1. The integral test for convergence.
Corollary to Theorem 9.5. The series 1
00
I"' n
n=l
known as the p-series, converges
if p >
1 and diverges
if p ~
1.
The proof of the Corollary is an immediate consequence of the integral test with f(x) = 1/xP. The details are left to the reader. For 0 < p ~ 1, the terms of the p-series tend to zero as n -+ oo although the series diverges. This fact shows that the converse of Theorem 9.1 is false. The hypothesis that un-+ 0 as n -+ oo does not imply the convergence of L~=l un. EXAMPLE.
Test the series 1
00
n~l (n + 2) log(n + 2) for convergence or divergence. Solution. We define f(x) = 1/(x + 2) log(x + 2) and observe that f is positive and nonincreasing with f(n) = 1/(n + 2) log(n + 2). We have
r
(x
+ 2)
~:g(x + 2) = f:+ u l~: u 2
=
fa+ 3
2
d(log u) log u
= log(log(a + 2)]
- log log 3.
Since log [log(a+ 2)]-+ +oo as a-+ +oo, the integral diverges. Therefore the 0 series does.
9.1. Tests for Convergence and Divergence
215
PROBLEMS
In each of Problems 1 through 10 test for convergence or divergence.
1
0()
t
2.
•=1P
1
0()
3.
ao
4
n+1
Ln·2"
• •=1
6.
~ 2n
ao
9.
1
L-2n + 3
•=1
•=1
+3
8 ~log n
--3-
0
n
•=1
•=1~ 1
0()
L -3 n
7. L.
L0()
Ln(n-+-2)
•=1
5.
1
0()
L-
L.
•=1
n
ao
L• •=1 e
10.
n3/2 nP
L -, p > 0 constant •=1
n!
11. Prove Theorem 9.4(a). [Hint: Use the fact that a bounded, nondecreasing sequence tends to a limit (Axiom of continuity).] 12. Prove Theorem 9.4(b).
L::'= For what values of p does the series L::'=
13. For what values of p does the series
1
log njnP converge?
14.
2
(log n)Pjn converge?
15. Prove the Corollary to Theorem 9.5. 16. Prove the Limit comparison theorem: Suppose that a. ~ 0, b. ~ 0, n = 1, 2, ... , and that
lim~=L>O.
n~oo
b,.
Then either L::'= 1 a. and L::'= 1 b. both converge on both diverge. [Hint. For sufficiently large n, we have tL < aJb. < !L. Now use the comparison test (Theorem 9.4) and the fact that the early terms of a series do not affect convergence.] 17. Use the result of Problem 16 to test for convergence: ao
L
•=1
2n 2 + n + 2 5n 3 + 3n ·
Take
2n 2 + n + 2 a = ----::::---.--------::,----5n 3 + 3n ' •
1 b.=-. n
18. Use the result of Problem 16 to test for convergence:
Take
a.= .,Yn2 + 5'
1 b.= liJ• n
216
9. Infinite Sequences and Infinite Series
9.2. Series of Positive and Negative Terms; Power Series When all the terms of a series are nonnegative, the Comparison test (Theorem
9.4) is a useful tool for testing the convergence or divergence of an infinite series. (See also the Limit comparison test in Problem 16 at the end of Section
9.1.) We now show that the same test may be used when a series has both positive and negative terms. Definition. A series L:'= 1 un which is such that L:'= 1 1unl converges is said to be absolutely convergent. However, ifL:'= 1 un converges and L:'= 1 lunl diverges, then the series L:'= 1 un is said to be conditionally convergent. The following theorem shows that if a series is absolutely convergent, then the series itself converges. Theorem 9.6. If L:'= 1 lunl converges, then L:'= 1 un converges and IJ1 Unl
~J
1
lunl·
PROOF. For n = 1, 2, ... , we define Vn
=
lunl
+ Un
2
Then we have and
0~
Vn
~ lunl,
0 ~ Wn ~ lunl·
Both L:'= 1 vn and L:'= 1 wn converge by the Comparison test (Theorem 9.4). Therefore L:'= 1 (vn - wn) = L:;, 1 un converges. Also,
~n~l Unl = IJ (vn- Wn)l ~ n~l (vn + wn) = Jl lunl· 1
D
Remark. A series may be conditionally convergent and not absolutely convergent. The next theorem shows that L:'= 1 ( -1r(1/n) is convergent. However, the series L:'= 1 1/n, a p-series with p = 1, is divergent; hence L:'= 1 ( -1r(1/n) is a conditionally convergent series. Theorem 9.7 (Alternating series theorem). Suppose that the numbers un, n = 1,
2, ... , satisfy the conditions: (i) the un are alternately positive and negative; (ii) lun+ll < lunl for every n; and (iii) limn->oo Un = 0.
217
9.2. Series of Positive and Negative Terms; Power Series
L:'=
Then 1 un is convergent. Furthermore, if the sum is denoted by s, then s lies between the partial sums sn and sn+l for each n. PROOF.
Assume that u 1 is positive. If not, consider the series beginning with
u2 , since discarding one term does not affect convergence. With u 1 > 0, we have clearly u 2 n-l
> 0 and u2 n < 0 for all n.
We now write
Since by (ii) above Iu2 k I < u 2 k-l for each k, each term in parenthesis is positive, and so s2 n increases with n. Also,
S2n =
U1 +
(u2
+ U3) +
(u4
+Us)+··· +
(u2n-2
+ U2n-1) +
U2n·
The terms in parentheses are negative and u2 n < 0. Therefore s2 n < u 1 for all n. Hence s2 n is a bounded, increasing sequence and therefore convergent to a number, say s; also, s2n ~ s for each n. By observing that s 2 n-l = s2n - u2n, we have s 2 n-l > s2 n for all n. In particular s 2 n-l > s 2 = u1 + u2 and so s 2 n-l is bounded from below. Also, S2n+l
=
S2n-1 +
(u2n
+ U2n+l)
1 or lun+l/unl-+ +oo, then there is an integer N such that lun+dunl > 1 for all n ~ N. Then lunl > iuNI for all n ~ N. Hence undoes not approach zero as n-+ oo. By Theorem 9.1 the series cannot converge. (iii) The p-series for all values of p yields the limit p = 1. Since the pseries converges for p > 1 and diverges for p ~ 1, the Ratio test can yield no information when p = 1. D EXAMPLE
1. Test for conditional and absolute convergence:
oo ( -1)nn L3n- . n=l
Solution. We use the Ratio test. Set un = ( -1)nn/3n. Then
Therefore
I 1-
1_ lim -Un+l ---p. n-+oo Un 3 The series converges absolutely.
D
The next theorem provides a useful test for many series.
Theorem 9.9 (Root test). Let L::"= 1 un be a series with either
Then (i) if p < 1, the series L::"= 1 Un converges absolutely; (ii) if p > 1, or if lunl 11n-+ +oo, the series diverges; (iii) if p = 1, the test gives no information. PROOF
(i) Suppose p < 1: choose e > 0 so small that p + e < 1 as well. Since lunl 11n-+ p, it follows that lunl 11n < p + e for all n ~ N if N is sufficiently large. Therefore lunl < (p + e)n for all n ~ N. We observe that 00
L (p + e)n
n=l
9.2. Series of Positive and Negative Terms; Power Series
219
is a convergent geometric series since p + e < 1. The Comparison test (Theorem 9.4) shows that L:'= 1 1unl converges. Hence (i) is established. (ii) Suppose p > 1 or lunl 11"-+ +oo. Choose e > 0 so small that (p- e)> 1. Therefore in Case (ii) p- e < lun1 1'" for all sufficiently large n. We conclude that n-+oo
and hence both L:'=t EXAMPLE
lunl and L:'=t u" are divergent series.
D
2. Test for convergence: 1
00
L
)". -(1 n=2 og n
Solution. We have
lunl 11"
1 log n
= ---+ 0
as
n-+ 00.
D
The series converges by the Root test. A power series is a series of the form
c0 + c 1 (x- a)+ c2 (x- af + ··· + c"(x- a)"+···, in which a and c;, i = 0, 1, 2, ... , are constants. If a particular value is given to x, then the above expression is an infinite series of numbers which can be examined for convergence or divergence. For those values of x in IR 1 which yield a convergent power series, a function is defined whose range is the actual sum of the series. Denoting this function by f, we write 00
f: x-+
L
c 0 (x - a)".
n=O
It will be established later that most of the elementary functions such as the trigonometric, logarithmic, and exponential functions have power series expansions. In fact, power series may be used for the definition of many of the functions studied thus far. For example, the function log x may be defined by a power series rather than by an integral, as in Section 5.3. If a power series definition is used, the various properties of functions, such as those given in Theorems 5.14 and 5.15 are usually more difficult to establish. We first state a lemma and then prove two theorems which establish the basic properties of power series.
Lemma 9.1. If the series L:'=t u" converges, then there is a number M such that lunl ~ M for all n. The proof is left to the reader. (See Problem 28 at the end of the Section.)
9. Infinite Sequences and Infinite Series
220
= x 1 where x 1 #a, then the series converges absolutely for all x such that lx- al < lx 1 - al. Furthermore, there is a number M such that
Theorem 9.10.Jf the series L~=o cn(x- a)n converges for x
lcn(x- a)nl:::;;
Mc~1 -=_ ~J
for
lx- al:::;; lx 1
-
and for all n.
al
(9.3) PROOF.
By Lemma 9.1 there is a number M such that
lcn(X 1
-
a)nl :::;; M
for all n.
Then (9.3) follows since
Ix
I
lx- aln (x- a)n ,n· t :::;; M I lcn(x- a)nl = lcn(Xl - atl· ( 1 -
a
x1
-
a
We deduce the convergence of the series at lx- al by comparison with the geometric series, the terms of which are the right side of the inequality in ~~
0
Theorem 9.11. Let L~=o cn(x - at be a given power series. Then either (i) the series converges only for x = a; or (ii) the series converges for all values of x; or (iii) there is a number R such that the series converges for lx- al < R and diverges for lx- al > R. PROOF. There are simple examples of series which show that (i) and (ii) may happen. To prove (iii), suppose there is a number x 1 #a for which the series converges and a number x 2 #a for which it diverges. By Theorem 9.10 we must have lx 1 - al :::;; lx 2 - al, for if lx 2 - al < lx 1 - al the series would converge for x = x 2 • Define the setS of real numbers
S = {p: the series converges for lx- al < p},
and denote R =supS. Now suppose lx'- al < R. Then there is apES such that lx'- al < p < R. By Theorem 9.10, the series converges for x = x'. Hence the series converges for all x such that lx- al < R. Now suppose that lx"- al = p' > R. If the series converges for x" then p' E Sand we contradict the fact that R = supS. Therefore the series diverges for lx- al > R, 0 completing the proof. EXAMPLE
3. Find the values of x for which the series oo (
n~l converges.
-1)n(x _ l)n 2nn 2
221
9.2. Series of Positive and Negative Terms; Power Series
Solution. We apply the Ratio test:
n Un+ll 1 IU:: =21x-11(n+ 1) 2
2
and
lun+11_11 . m l1 -- --Xn-+oo
Un
2
11 .
Therefore the series converges for tlx- 11 < 1 or for -1 < x < 3. By noting that for x = -1 and x = 3, the series is a p-series 1 with p = 2, we conclude that the series converges for all x in the interval - 1 ::::; x ::::; 3 and it diverges 0 for all other values of x. PROBLEMS
In each of Problems 1 through 16, test the series for convergence or divergence. If the series is convergent, determine whether it is absolutely or conditionally convergent. 00
1.
3.
L
(
n.
•=1
oo ( -1)"- 1 (n- 1)
I
n +1
4.
-1 )"(4/3)"
6.
----,2.--------:---
00
I
(
n
•=1
7.
4
~(-1)"(n+1) L....
oo ( -1)"(2n 2
I
Ioo 00
I
•=1
n+ 1
(-
2)"
-3-
n
I (-1)"r"
n=l
-----:-c:-::-:-::---
·=1
(10)"
oo
(-1r 1
I -:-----=-= n(n - 1/2)
•=1
I
(-1)•-1n!
•= 1
1· 3 · 5 · · · (2n- 1)
oo (-1)"·2·4-6-··2n
I
-:-------c---=-
1-4- 7 · · · (3n - 2) (-1)"+1
00
10.
I (n--,..-----,+ 1) log(n + 1)
•=1
+ 1)
-'----'-----------=:-'---------'-
00
15.
3n + 2)
( -1)"+ 1 log(n
·=1
13.
3
I
•=1
n
•=1
11.
8.
v/n
n=l
9.
2.
I
•=1
5.
oo ( -1)•-1n!
-1)"- 1 (10)"
00
12.
I
(
-1r 1 log n n
•=1
14.
2
oo n' I,; e
•=1
16. Ioo(-n-)" 2n + 1 •=1
In each of Problems 17 through 24, find all the values of x for which the given power series converges. 00
17. L(n+1)x" n=O 1
00
18.
(x - 2)"
I-n=l
v/n
Actually, for x = 3 the absolute values of the terms form a p-series.
9. Infinite Sequences and Infinite Series
222
-I :(3/2)"x" n+1 00
19.
00
2t
23 _
00
2o.
.~o
n!(x - 3)" 1. 3 ... (2n _ 1)
I:
-:--::---'----:-=-----'----:-:-
f
log(n
.~,
oo
+ 2n + 1)x" 3 2"(n + 1)
.f-o 00
24.
+ 2)"
(2n 2
22. "'
+ 1)2"(x + 1)" n+1
.~,
n(x
I: ------=-=------2"
.~,
I:
(
-1
n)2"x" r' (log . 2
3n
.~,
u.
25. Ifi;::"~ 1 is a convergent series of positive terms, show that I:::"~ 1 for every p > 1.
u: is convergent
26. Find the interval of convergence of the binomial series ~
1 + L.
.~,
m(m-1) ... (m-n+l) x" n!
m a constant.
*27. Let I;::"~ 1 u. be a conditionally convergent series with terms satisfying the conditions of Theorem 9.7. Let A be any real number. Show that by rearranging the terms of the series, the sum will be a number in the interval (A - 1, A + 1). 28. Prove Lemma 9.1. [Hint: Use Theorem 9.1 and the fact that a finite number of terms has (at least) one which is largest in absolute value.] 29. Give examples of Cases (i) and (ii) in Theorem 9.11.
9.3. Uniform Convergence of Sequences Let {!.} be a sequence of functions with each function having a domain containing an interval I of IR 1 and with range in IR 1 . The convergence of such a sequence may be examined at each value x in I. The concept of uniform convergence, one which determines the nature of the convergence of the sequence for all x in I, has many applications in analysis. Of special interest is Theorem 9.13 which states that if all the f. are continuous then the limit function must be also. Definition. We say the sequence {f.} converges uniformly on the interval I to the function f if and only if for each 6 > 0 there is a number N independent of x such that lf.(x) - f(x)i
N.
(9.4)
Uniform convergence differs from ordinary pointwise convergence in that the integer N does not depend on x, although naturally it depends on 6. The geometric meaning of uniform convergence is illustrated in Figure 9.2. Condition (9.4) states that if 6 is any positive number, then for n > N the graph of y = f.(x) lies entirely below the graph of f(x) + 6 and entirely above the graph of f(x) - 6.
223
9.3. Uniform Convergence of Sequences y y=f(x)+e
y =f.(x) y = f(x)
y =f(x)-e
0
Figure 9.2. Illustrating uniform convergence.
It can happen that a sequence {.f..(x)} converges to f(x) for each x on an interval I but that the convergence is not uniform. For example, consider the functions
l={x:O:::;x:::;l}. The graphs of fn for n = 1, 2, 3, 4 are shown in Figure 9.3. If x =I 0, we write
2x/n
fix)
= x2 + (1/n2)
--~----------------------~L----------+
0
Figure 9.3. Illustrating nonuniform convergence to f(x)
=0.
X
224
9. Infinite Sequences and Infinite Series
and observe that fn(x) -+ 0 for each x > 0. Moreover, f,.(O) = 0 for all n. Hence, setting f(x) 0 on I, we conclude that fn(x)-+ f(x) for all x on I. Taking the derivative off,., we find
=
, f,.(x)
=
2n(1 - n2x 2) (1 + n2x2)2 .
Therefore fn has a maximum at x = 1/n with f,.(1/n) = 1. Thus if e < 1, there is no number N such that 1/n(x)- f(x)l < e for all n >Nand all x on I; in particular 1/n(l/n) - /(1/n)l = 1 for all n. The definition of uniform convergence is seldom a practical method for deciding whether or not a specific sequence converges uniformly. The next theorem gives a useful and simple criterion for uniform convergence.
Theorem 9.12. Suppose that fn, n = 1, 2, 3 ... , and f are continuous on I = {x: a~ x ~ b}. Then the sequence fn converges uniformly to f on I if and only if the maximum value en of lfn(x) - f(x)l converges to zero as n-+ oo. PROOF
(a) First, suppose the convergence is uniform. Let e > 0 be given. Then there is an N such that lf,.(x) - f(x)l < e for all n > N and all x on I. Since lf,.(x) - f(x)l is continuous on I, it takes on its maximum value at some point Xn. Then en = 1/n(xn) - f(xn)l. Hence en < e for all n > N. Since there is an N for each e > 0, it follows that en-+ 0 as n-+ oo. (b) Suppose that en-+ 0 as n-+ oo. Then for each e > 0 there is anN such that en < e for all n > N. But, then, since en is the maximum of lf,.(x)- f(x)l, 0 we have 1/n(x) - f(x)l ~ en < e for all n > N and all x on I. EXAMPLE.
Given the sequence n = 1, 2, ... ,
show that f,.(x)-+ 0 for each x on I = { x: 0 or not the convergence is uniform.
~
x
~
1} and determine whether
Solution. Since f,.(O) = 0 for each n, we have fn(O)-+ 0 as n-+ oo. For x > 0, we divide numerator and denominator by n2 and find fn(x) = x2
x/Jn + (1jn2)'
and it is evident that f,.(x)-+ 0 as n-+ oo. Taking the derivative, we obtain ,
f,. (x) =
n312( 1
(1
_ n2x2)
+ n2x2)2
It is clear that J:(x) = 0 for x = 1/n; also fn(ljn) = Jn/2. Therefore en =
225
9.3. Uniform Convergence of Sequences
fn(l/n) does not converge to 0 as n --+ oo and hence the convergence of fn is not
D
uniform.
Although Theorem 9.12 is useful it cannot be applied unless the limit function f is known. The importance of uniform convergence with regard to continuous functions is illustrated in the next theorem.
Theorem 9.13. Suppose that !,, n = 1, 2, ... , is a sequence of continuous functions on an interval I and that is continuous on I. PROOF.
{!,} converges uniformly to f on I. Then f
Suppose e > 0 is given. Then there is an N such that 1/,(x) - f(x)l
N.
and all
(9.5)
Let x 0 be any point in I. Since fN+l is continuous on I, there is a lJ > 0 such that lfN+l (x)- fN+l (x 0 )1
e
< 3 for all x
E
I
lx- x 0 1< /J.
such that
(9.6)
Also, by means of (9.5) and (9.6) we find lf(x) - f(xo)l ::::; lf(x) -
fN+l (x)l
+ lfN+l (x) e
fN+l (xo)l
e
e
+ lfN+l(xo)- f(xo)l < 3 + 3 + 3 = e, which is valid for all x on I such that lx- x 0 1< /J. Thus x 0 , an arbitrary point of I.
f
is continuous at D
The next result shows that a uniformly convergent sequence of continuous functions may be integrated term-by-term.
Theorem 9.14 (Integration of uniformly convergent sequences). Suppose that each fn, n = 1, 2, ... , is continuous on the bounded interval I and that {fn} converges uniformly to f on I. Let c E I and define Fn(X) =
lx
fn(t) dt.
Then f is continuous on I and Fn converges uniformly to the function F(x) =
lx
f(t) dt.
PROOF. That f is continuous on I follows from Theorem 9.13. Let L be the length of I. For any e > 0 it follows from the uniform convergence of Un} that
226
9. Infinite Sequences and Infinite Series
there is an N such that
l.f..(t) - f(t)l
8
N
for all
We conclude that
IFn(x) - F(x)l =
If'
and all t on
[.f..(t) - f(t)] dt
~ 1f." l.f..(t) -
I.
I
f(t)l dt 1
8
~-lx-cl ~8
L
for all n >Nand all x e I. Hence {Fn} converges uniformly on I.
0
The next theorem illustrates when we can draw conclusions about the term-by-term differentiation of convergent sequences.
Theorem 9.15. Suppose that {f..} is a sequence of functions each having one continuous derivative on an open interval I. Suppose that .f..(x) converges to f(x) for each x on I and that the sequence J: converges uniformly tog on I. Then g is continuous on I and f'(x) = g(x) for all x on I. PROOF. That g is continuous on I follows from Theorem 9.13. Let c be any point on I. For each n and each x on I, we have
f." f:(t) dt = .f..(x) -
.f..(c).
Since {!:} converges uniformly tog and .f..(x) converges to f(x) for each x on I, we may apply Theorem 9.14 to get
f." g(t) dt = f(x)- f(c). The result follows by differentiating (9. 7).
(9.7)
0
Many of the results on uniform convergence of sequences of functions from R 1 to R 1 generalize directly to functions defined on a set A in a metric space Sand with range in R 1 •
Definition. Let A be a set in a metric space S and suppose that f..: S-+ R 1, n = 1, 2, ... , is a sequence offunctions. The sequence {f..} converges uniformly on A to a function f: A -+ R1 if and only if for every 8 > 0 there is a number N such that l.f..(x) - f(x)l < 8 for all x e A and all n > N. The next theorem is an extension of Theorem 9.13.
9.3. Uniform Convergence of Sequences
227
Theorem 9.16. Suppose that each f, is continuous on a set A in a metric space S where f,: S-+ IR 1 • If {f,} converges uniformly to f on A, then f is continuous on A.
The proof is similar to the proof of Theorem 9.13 and is left to the reader. Theorem 9.14 has a generalization to functions defined inN-dimensional Euclidean space.
Theorem 9.17. Let F be a closed figure in IRN and suppose that f,: F-+ ~Rl, n = 1, 2, ... , is a sequence of continuous functions which converges uniformly to f on the figure F. Then f is continuous on F and
f
f dVN = lim
F
n-ex>
r f, dVN.
JF
The proofis similar to the proof of Theorem 9.14 and is left to the reader. Let {f,} be a sequence defined on a bounded open set Gin IRN with range in IR 1• Writingfn(x 1 , x 2 , ••• , xN) for the value off, at(x 1 , x 2 , ••• , xN), we recall that
are symbols for the partial derivative off, with respect to xk.
Theorem 9.18. Let k be an integer such that 1 ~ k ~ N. Suppose that for each n, the functions f,: G-+ IR 1 and fn.k are continuous on G, a bounded open set in IRN. Suppose that Un(x)} converges to f(x) for each x e G and that Un.d converges uniformly to a function g on G. Then g is continuous on G and fk(x) = g(x)
x e G.
for all
The proof is almost identical to the proof of Theorem 9.15 and is omitted. If a sequence of continuous functions {fn} converges at every point to a continuous function f, it is not necessarily true that
f f f,dV-+
f dV
as
n -+ oo.
Simple convergence at every point is not sufficient as the following example shows. We form the sequence (for n = 2, 3, 4, ... )
0
1
~X~-,
n 1 2 -~X~-, n n
f,: X-+ 0,
2
-~x~l.
n
9. Infinite Sequences and Infinite Series
228 y
n=4
4
/ 4 (x)
n=3
3
j,(x)
n=2
2
0
2
Figure 9.4. f. converges to f but J~ f. does not converge tog f.
It is easy to see (Figure 9.4) that f.(x)
{x: 0::::;; x::::;; 1}, but that 1=
I
1
f,(x) dx
-+
L
+
1
f(x)
f(x) dx
=0
for each x
E
I
=
= 0.
Theorems 9.16, 9.17, and 9.18 may be generalized to sequences of functions with domain in a metric space (S1 , dd and range in another metric space (S2 , d2 ). See Problem 19 at the end of this section. PROBLEMS
In each of Problems 1 through 10 show that the sequence {f,(x)} converges to f(x) for each x on I and determine whether or not the convergence is uniform. 2x l.f.:x-+--, f(x):=O, 1 + nx 2.
f.: X-+
3.
f.:
cos nx
Jn ,
X-+ -1
n3 x 4- , +nx -
f(x) := 0,
f(x) := 0,
I={x:O~x~1}.
I= {x: 0 ~X~
1}.
I= {x: 0 ~X~ 1}.
9.3. Uniform Convergence of Sequences
4. f.:x-+
n3 x 4
1+n x
2,
nx 2 5.f.:x-+--, 1 + nx
1
6. f.: x-+ ;: ..,;x
f(x)::O,
I={x:a.;;;x< oo,a>O}.
I={x:O.;;;x.;;;1}.
f(x)=x,
1
+ -cos(1/nx),
f(x) =
n
sin nx 7. f.: x-+ - - , 2nx
f(x)
8. f.: x-+ x"(1- x)Jn,
1 - x•
9. f.: X-+--,
1-x
10. f.: x-+ nxe-""'\
=0, f(x)
=0,
1/Jx,
I= {x: 0 < x,:;;; 2} .
I= {x: 0 < x < oo }.
= 0,
1 -, f(x)=1-x f(x)
229
I= {x: 0,:;;; x.:;; 1}.
I={x:-~.;;;x.;;;~}· 2 2
I= {x: 0,:;;; x,:;;; 1}.
11. Show that the sequence f.: x-+ x" converges for each x that the convergence is not uniform.
E
I = {x: 0 ,:;;; x ,:;;; 1} but
12. Given that J.(x) = (n + 2)(n + 1)x"(1 - x) and that f(x) = 0 for x on I= {x: 0 ,:;;; x ,:;;; 1}. Show that f.(x)-+ f(x) as n-+ oo for each x E I. Determine whether or not J~J.(x) dx-+ J5f(x) dx as n-+ oo.
13. Prove Theorem 9.16. 14. Prove Theorem 9.17. 15. Prove Theorem 9.18. 16. Give an example of a sequence of functions {f.} defined on the set A = { (x, y): 0 ,:;;; x .:;; 1, 0 .:;; y ,:;;; 1} such that f. converges to a function fat each point (x, y) E A but JAJ.(x, y) dV JAf(x, y) dV.
+
17. Suppose that {f.} converges uniformly to f and {g.} converges uniformly tog on a set A in a metric space S. Show that {f. + g.} converges uniformly to f + g. 18. (a) Suppose {!.} and {g.} are bounded sequences each of which converges uniformly on a set A in a metric space S to functions f and g, respectively. Show that the sequence {f.g.} converges uniformly to fg on A. (b) Give an example of sequences {f.} and {g.} which converge uniformly but are such that {f.g.} does not converge uniformly. 19. Formulate a definition of uniform convergence of a sequence {!.} of mappings from a set A in a metric space (S1 , dd into a metric space (S2 , d 2 ). Prove that if {f.} are continuous and converge uniformly to a mapping f, then f is continuous. 20. Prove the following generalization of Theorem 9.17. Suppose that F is a figure in IRN and that f.: F-+ IR 1 is integrable on F for n = 1, 2, ... If {f.} converges uniformly to f on F, then f is integrable over F and
r fdVN =lim Jr f.dVN.
JF
[Hint: First prove that f is integrable.]
n-oo
F
9. Infinite Sequences and Infinite Series
230
9.4. Uniform Convergence of Series; Power Series Let uk(x), k = 1, 2, ... , be functions defined on a set A in a metric spaceS with range in IR 1 . Definition. The infinite series L;;"= 1 uk(x) has the partial sums sn(x) = Lk=l uk(x). The series is said to converge uniformly on a set A to a functions if and only if the sequence of partial sums {sn} converges uniformly to son the set A. The above definition shows that theorems on uniform convergence of infinite series may be reduced to corresponding results for uniform convergence of sequences. Theorem 9.19 (Analog of Theorem 9.13). Suppose that un, n = 1, 2, ... , are continuous on a set A in a metric spaceS and that L::"= 1 un(x) converges uniformly on A to a function s(x). Then s is continuous on A. The next theorem is the analog for series of a generalization of Theorem 9.14. In this connection see Problem 20 of Section 9.3.
Theorem 9.20 (Term-by-term integration of infinite series) (a) Let un(x), n = 1, 2, ... , be functions whose domain is a bounded interval I in IR 1 with range in IR 1 • Suppose that each un is integrable on I and that 1 un(x) converges uniformly on I to s(x). Then s is integrable on I. If c is in I and Un, S are defined by
L::"=
Un(x) =
Lx un(t) dt,
S(x)
=
Lx s(t) dt,
then L::"= 1 Un(x) converges uniformly to S(x) on I. (b) (See Theorem 9.17.) Let un(x), n = 1, 2, ... , be defined on a figure Fin IRN with range in IR 1 . Suppose that each un is integrable on F and that L::"= 1 un(x) converges uniformly on F to s(x). Then sis integrable on F and
fI
F
n=l
Un
dVN =
I
F
S
dVN.
The next result is the analog for series of Theorem 9.18. Theorem 9.21 (Term-by-term differentiation of infinite series). Let k be an integer with 1 ~ k ~ N. Suppose that un and un,k• n = 1, 2, ... , are continuous functions defined on an open set G in IRN (range in IR 1 ). Suppose that the series L::"= 1 un(x) converges for each x E G to s(x) and that the series L~ 1 un,k(x) converges uniformly on G to t(x). Then s,k(x)
=
t(x)
for all x in G.
9.4. Uniform Convergence of Series; Power Series
231
The proofs of Theorems 9.19, 9.20, and 9.21 are similar to the proofs ofthe corresponding theorems for infinite sequences. The following theorem gives a useful indirect test for uniform convergence. It is important to observe that the test can be applied without any knowledge of the sum of the series.
Theorem 9.22 (Weierstrass M-test). Let un(x), n = 1, 2, ... , be defined on a set A in a metric spaceS with range in IR 1 • Suppose that lun(x)l ~ Mn for all nand for all x e A. Suppose that the series of constants L:'= 1 Mn converges. Then L:'= 1 un(x) and L:'=1 lun(x)l converge uniformly on A. PRooF. By the Comparison test (Theorem 9.4) we know that L:'= 1 1un(x)l converges for each x. Set n
tn(x) =
oo
L luk(x)l k=l
and
L luk(x)l.
t(x) =
k=l
From Theorem 9.6 it follows that L:'= 1 un(x) converges. Set n
sn(x) =
L uk(x),
k=l
00
s(x) =
L uk(x).
k=l
Then
00
~
L luk(x)l = k=n+l
lt(x)- tn(x)l
We define s = Lk=l Mk> sn = I;;=l Mk. Then Lk=n+l Mk = s - Sn; since S - Sn -+ 0 as n -+ oo independently of x, we conclude that the convergence of {sn} and {tn} are uniform. D The next theorem on the uniform convergence of power series is a direct consequence ofthe Weierstrass M-test.
Theorem 9.23. Suppose that the series L:'=o cn(x - a)" converges for x = x 1 with x 1 =F a. Then the series converges uniformly on I = { x: a - h ~ x ~ a + h} for each h < lx 1 - al. Also, there is a number M such that lcn(x-
a)" I ~ M · (lx ~ aY 1
(9.8)
for lx- al ~ h < lx 1
-
al.
PROOF. Inequality (9.8) is a direct consequence of the inequality stated in Theorem 9.10. The series L:'=o M(h/lx 1 - al)" is a geometric series of con-
9. Infinite Sequences and Infinite Series
232
stants which converges. Therefore the uniform convergence of'L~=o cix- at D follows from the Weierstrass M-test.
Remarks (i) The number h in Theorem 9.23 must be strictly less than lx 1 - ai in order that the series converge uniformly on I. To see this consider the series oo
(-1)nxn
n=l
n
I~~
which converges for x = 1 but diverges for x = -1. If this series were to converge uniformly for lxl < 1, then it would converge uniformly for lxl :::; 1. (The reader may establish this fact.) (ii) If the series in Theorem 9.23 converges absolutely for x = Xt. then we may choose h = lx 1 - ai and the series converges uniformly on I= {x: lx- ai:::; lx 1 - aj}. To see this we observe that for lx- ai :::; lx 1 - ai, we have
EXAMPLE
1. Given the series 00
L
(n
+ 1)xn,
(9.9)
n=O
find all values of h such that the series converges uniformly on I
=
{x: Ixi :::; h}.
Solution, For lxl :::; h, we have l(n + 1)xnl :::; (n + 1)hn. By the Ratio test, the series L~=o (n + l)hn converges when h < 1. Therefore the series (9.9) converges uniformly on I= {x: lxl :::; h} if h < 1. The series (9.9) does not converge for x = ± 1, and hence there is uniform convergence if and only if h < 1.
D EXAMPLE
2. Given series (9.10)
findallvaluesofhsuchthatthes eriesconvergesuniformlyoni
=
{x: lxl:::; h}.
Solution. For lxl :::; h, we have lxn/n 2 1 :::; hn/n 2 • The p-series L~= 1 1/n 2 converges and, by the Comparison test, the series L~=l hnjn 2 converges if h:::; 1. By the Ratio test, the series (9.1 0) diverges if x > 1. We conclude that the series D (9.10) converges uniformly on I= {x: lxl:::; 1}. Lemma 9.2. Suppose that the series 00
f(x)
=
L
cn(x - a)n
(9.11)
n=O
converges for ix- ai < R with R > 0. Then f and
f' are continuous on
9.4. Uniform Convergence of Series; Power Series
233
lx- a!< Rand f'(x) =
co
L ncn{x -
a)"- 1 for
n=1
lx - a! < R.
(9.12)
PROOF. Choose x 1 such that lx 1 - al < R and then choose h so that 0 < h < lx 1 - a!. By Theorem 9.23, the series (9.11) converges uniformly on I= {x: lx- al ~ h}, and there is a number M such that
lc,.(x - a)" I
~ M Cx ~ a!)" 1
for
lx - a!
~ h.
According to Theorem 9.19, the function f is continuous on I. Also, we find
Inc, (x - a)n-11
~
nM(lxt h ai)" = u,..
n Ic, lh"-1 ~ h
~
From the Ratio test, the series L::"= 1 u, converges, so that the series (9.12) converges uniformly on I. Hence by Theorem 9.19, the function f' is continuous on I. Since h may be chosen as any positive number less than R, we conclude that f and f' are continuous for lx- a! < R. 0 With the aid of the above lemma, we obtain the following theorem on term-by-term differentiation and integration of power series.
Theorem 9.24. Let f be given by co
f(x) =
L
n=O
c,.(x - a)" for
lx - a! < R with R > 0.
(9.13)
(i) Then f possesses derivatives of all orders. For each positive integer m, the derivative pm>(x) is given for lx- a! < R by the term-by-term differentiation of (9.13) m times. (ii) IfF is defined for lx - a! < R by F(x) =
f.."' f(t) dt,
then F is given by the series obtained by term-by-term integration of the series (9.13). (iii) The constants c,. are given by J!">(a)
c,.=--,-. n.
(9.14)
PRooF. The proof of(i) is obtained by induction using Lemma 9.2. We obtain (ii) by application of Theorem 9.20. To establish (iii) differentiate the series (9.13) n times and set x =a in the resulting expression for j if 0
X -:1-
if X
0,
= 0.
By differentiation, we find that for x -:1- 0
f'(x)
= 2x-Je-ltx>,
f"(x) = x- 6 (4 - 6x 2 )e-lfx>,
f, where Pn is a polynomial. By l'Hopital's rule, it is not hard to verify that J(O) = 0 for every n, and so f has continuous derivatives of all orders in a neighborhood of 0. The series (9.15) for f is identically zero, but the function f is not, and therefore the power series with a = 0 does not represent the function.
f: I-+ IR 1
be infinitely differentiable at a point a e I and suppose that the series (9.15) has a positive radius of convergence. Then f is said to be analytic at a. A function f is analytic on a domain if and only if it is analytic at each point of its domain.
Definitions. Let
A function must have properties in addition to infinite differentiability in order to be representable by a power series and therefore analytic. The principal tool for establishing the validity of power series expansions is given by Taylor's theorem with remainder.
Theorem 9.25 (Taylor's theorem with remainder). Suppose that f and its first n derivatives are continuous on an interval containing I= {x: a:::;;; x:::;;; b}. Suppose that f(n+ll(x) exists for each X between a and b. Then there is a with a < < b such that
e
e
n
f(b)
=
fUl(a)
.
.L -.J.1-(b- a)'+ Rn,
]=0
235
9.4. Uniform Convergence of Series; Power Series
where (9.16) PROOF.
We define Rn by the equation n
f(b) =
J'-il(a)
L -. 1-(b j=O J.
a)l
+ Rn,
that is,
Rn
=f(b) - f. JU~(a) (b j!
a)l.
j=O
We wish to find the form R" takes. For this purpose define; for x on I by the formula n JW(x) (b _ x)n+1 ,P(x) = f(b)- L -.1-(b- x)l- R"(b )"+1 · J=O 1· -a Then; is continuous on I and ,P'(x) exists for each x between a and b. A simple calculation shows that ,P(a) = ,P(b) = 0. By Rolle's theorem there is a number with a< < b such that ,P'(e) = 0. We compute
e
e
,P'(x) = -f'(x) +
n
J a, then a < then X < < a and e~ < ea. Hence
e
lx - aln+l Rn(a, x) ~ C(a, x) ( 1) 1 n+.
where
e
0 there is an integer n such that IL~=l an- sl < e for every integer m > n. We might compare the set 1, 2, ... , n to S' and the set 1, 2, ... , m to S". However, in defining the sum over a set S, nothing is said about how the finite set S' is chosen; it is not necessarily the first n integers if, say, S is the collection of natural numbers. The definition given for a sum overS allows us to choose a finite set of elements arbitrarily in the collectionS. For this reason we refer to (9.23) as an unordered sum. The next theorem shows that the sum over an infinite set can have at most one values. Theorem 9.32. Let S be an irifinite set and suppose that f: S--+ IR 1 is a given function. Then f can have at most one sum s over S.
9. Infinite Sequences and Infinite Series
244
Assume thatfhas two sums s 1, s 2 overS with s 1 < s 2 • We shall reach a contradiction. We set e = !(s 2 - s 1). Then there are finite sets S~, s; such that PROOF.
~p~" f(p)- s1 < e
and
1
for all finite sets S", S" such that S~ which contains S~ us;. Then
S" and s;
c
L f(p) < s 1 + e
peS*
IJ;./(P)-
and
for this set S*. We conclude that s 2 definition of e, is impossible.
-
s2
c
S". LetS* be any finite set
e
0 there is a finite set S~ c: S0 such that
IpeSL .. f(p)-sl<e 0
for every finite setS~ containing S~. We denote SJ. = by the definition of g, we have
T(S~),
sr =
T(S~).
Then
IqeSL . g(q)- sl < e 1
for every finite set S~ containing s;,. Hence Lqes, g(q) = s.
D
We recall that the symbol N is used to denote the set of all positive integers. We also use the symbol N 0 to designate the set of all nonnegative integers, i.e. N 0 = N u {0}.
Theorem 9.38. Let f: N-+ IR 1 be given. For each n EN, we let an= f(n). Then f has a sum over N if and only if L~=l an converges absolutely. Also, the sums are equal: 00 L an= L an. n=l
ne N
Suppose thatfhas a sum over N. Then lfl,J+, andf- also have sums over N. Lets+ = Lne N a: where a: =!(I ani +an), and suppose that e > 0 is given. Then there is a finite set N' such that PROOF.
s+ - e
n=l (m,n)eSn'
=
1
n'
L2 n=l n
>I-. n=l n Thus, if M is any positive number, we can choose n' so large that
L
(m,n)eSn'
F(m, n) > M.
Therefore F does not have a sum overS and, by the Comparison test, neither
D
~~
The next two theorems are almost direct consequences of results already established. We leave the details of the proofs to the reader. Theorem 9.40. Iff: S-+ IR 1 and g: S-+ IR 1 have sums over S, and f(p) ~ g(p) for all p e S, then g(p). f(p) ~
I
peS
I
peS
252
9. Infinite Sequences and Infinite Series
Theorem 9.41 (Sandwiching theorem). Let f: S--+ lll 1 , g: S--+ lll 1 , and h: S--+ lll 1 be given. Suppose that f and h each has a sum overS and that f(p) ~ g(p) ~ h(p) for all p e S. Then g has a sum over S. Although the sets S may be quite arbitrary, we now show that iff has a sum over S then the set of points of S where f does not vanish is severely restricted.
Theorem 9.42. Suppose that f: S --+ lll 1 has a sum over S. Let S* = {p e S: f(p) =I 0}. Then S* is countable. PROOF.
For n = 2, 3, ... , define Sn =
{p
E
S:
~ ~ lf(p)l < n ~ 1}.
The reader may complete the proof by showing that S* each S" must be a finite set.
=
U:'=
2
Sn and that D
Let A be a set in a metric space T and suppose that S is an arbitrary set. Let f be defined for x e A and peS with values in lll 1 ; that is, f: A x S--+ lll 1• We write f(x; p) for the function values off For each x in A, the function f may or may not have a sum over S; when f has a sum, we denote it by s(x). For unordered sums we make the following definition, one which corresponds to uniform convergence for infinite series.
Definition. Let f: A x S--+ lll 1 be given. We say that f has a sum uniformly on A over S if and only iff has a sum over S for each x e A and, for every 8 > 0 there is a finite set S' contained in S such that
~P~" f(x; p)- s(x)l
0 there is a positive integer P such that lumn- al < 8 for all (m, n) such that m > P and n > P. We write {umn}-+ a as (m, n)-+ oo. In analogy with the formal definition of an infinite series given in Section 9.1, we define a double series formally as an ordered pair ({umn}, {smn}) of double sequences such that n
m
Smn
L uij. L j=O
=
i=O
The umn• m, n = 0, 1, 2, ... , are called the terms of the double series ( {umn}, {smn} ), and the smn are called its partial sums. The double series is said to be convergent if and only if there is a numbers such that Smn-+ s as (m, n)-+ oo. Otherwise, the double series is divergent. We will most often write a double series in the form 00
L
uii
i,j=O
rather than use the ordered pair symbol. The following theorem is a direct consequence of the definition of convergence and the theorems on limits.
Theorem 9.47 (a) If the double series L::.n=o umn• L::;',n=o vmn are both convergent and c and d are any numbers, then the series L::.n=o (cumn + dvmn) is convergent and 00
00
00
L
(cumn
+ dvmn) = c
m,n=O
L
umn
m,n=O
(b) If the series L::;',n=o Umn is divergent and divergent.
+d
L
Vmn·
m,n=O
if c =I= 0, then L::;',n=o cumn is
The next theorem shows the relationship between the convergence of a double series and an unordered sum.
Theorem 9.48. Suppose that the terms umn of a double sequence u: No X No-+ IR 1 are all nonnegative. Then the double series L::.n=o umn is convergent if and only if the function u has a sum overS= N 0 x N 0 • Furthermore, 00
L Umn m,n=O
=
L umn· (m,n)eS
(9.29)
9. Infinite Sequences and Infinite Series
256 PROOF
(a) Suppose that u has a sum over S which we denote by s. Let e > 0 be given. Then there is a finite subset S' of S such that Umn>S-6. L (m,n)eS'
Let P be a positive integer such that P > m and P > n for all (m, n) E S'. Since 0 for all i, j, it follows that
uii ~
s-e
0 be given. Then there is a positive integer P such that P ~ m and P ~ n for (m, n) E S', and furthermore the inequality lsmn- sl < e holds whenever m ~ P and n ~ P. Hence p
p
Umn ~ L L Umn = spp ~ s L m=l n=l (m,n)eS'
+ e.
Since e and S' are arbitrary, u has a sum over S and its value is s.
D
It is clear that corresponding to Theorems 9.47 and 9.48 there are theorems for triple series, quadruple series and, in fact, for multiple series of any order.
Theorem 9.49. Let u: N0 x N0 -+ IR 1 be given and suppose that its terms are Umn· Then the double series L~.n=O Umn is absolutely convergent if and only if the function u has a sum over S = N 0 x N 0 • In the case of convergence, Equation (9.29) holds. The above theorem is a direct consequence of Theorem 9.48 and the related theorems on unordered sums. We leave the details of the proof to the reader. Theorems 9.48 and 9.49 make the theory of absolutely convergent double series a special case of the theory of unordered sums given in Section 9.6. The next result is a special case of a general theorem on unordered sums (Theorem 9.51 below). -+ IR 1 be given with its terms denoted by umn· Suppose that L~.n=o umn is absolutely convergent. Then the single series
Theorem 9.50. Let u: N0 x N0
is absolutely convergent for each m = 0, 1, 2, ... , and the series L~=o Umn is
9.7. Multiple Sequences and Series
257
absolutely convergent for each n = 0, 1, 2, .... If we define 00
v, = L
W,
umn•
m=O
=
L
Umn
m+n=r
then the series L:::=o Um, L;:'=o V,, L~o W, are all absolutely convergent and have the same value as L;::,n=o Umn· We first show that L;:'=o umn is convergent for each m when umn ~ 0. By Theorem 9.49, we know that u has a sum overS= N 0 x N 0 • Since the partial sum L~=o umn is a finite subset of S, these partial sums are uniformly bounded. Hence the series is convergent for each m. Now consider the partial sum L;:=o Um. For any e > 0 and any positive number M, there is an integer P such that PROOF.
p
00
L Umn L Umn > n=O n=O
ejM.
Hence for any positive integer M, it follows that M
oo
M
M
P
L L Umn < L L Umn + e. um = m=O m=O n=O n=O m=O L
Since u has a sum overS, we conclude that the series L:::=o Um converges. Now for any e > 0 there is a finite set S' c S such that
L
umn>
L
Umn-e.
(m,n)eS'
(m,n)eS
Also, there is an integer M' which we can take larger than the greatest value of min S', such that oo
um > m=O L
M'
M'
oo
L
L
L
um = m=O
umn >
m=O n=O
L
Umn -
e.
(m,n)eS'
Since e is arbitrary it follows that L:::=o um = L:::.n=O Umn· We treat the case where umn may be positive or negative by using the hypothesis of absolute convergence and separating the terms of the series into positive and negative parts. As in previous proofs, we apply the above argument to each of the series 0 obtained in this way. The proofs for L;:'=o V,, L~o W, are similar. Theorem 9.50 is a special case of a general theorem on unordered sums which may be considered as a generalized associative law. This result allows us to extend the above theorem to multiple series of any order.
Theorem 9.51. Let S' be a set. To every x in S' we associate a set denoted Sx. Define S = {(x, y): xES', y E Sx}· Suppose that f: S--+ ~ 1 has a sum overS. Then f has a sum over Sxfor each xES'. If we define g(x) = Lyesxf(x, y), then g has a sum over S' and
L g(x) = L [ L f(x, y)J = xeS' yeSx
xeS'
L (x,y)eS
f(x, y).
258
9. Infinite Sequences and Infinite Series
The proof follows the pattern of the proof of Theorem 9.50, and we leave the details to the reader. We illustrate Theorem 9.51 with an example in triple series. LetS' = N x N and for each xeS', let S"' = N 0 • Then S = N x N x N 0 • Suppose f: S-+ R1 has a sum overS; we denote the terms off by u1mn• (I, m, n) e S. Theorem 9.51 then shows that g(x)
=L
00
f(x, y) =
yeS,
L Ulmn n=O
is convergent for all (I, m) e S'. Also, if we denote the terms of g(x) by v1m, (I, m) e S', then 00
00
00
L v,m = l,m=l L n=O L u,""' = (l,m,n)eS L Ulmn· l,m=l The next theorem is a partial converse of Theorem 9.51.
Theorem 9.52. Suppose that S', S, and S"' are as in Theorem 9.51. Let f: S-+ R1 be given. Suppose that f has a sum over S"' for each xeS'. Define g(x) = Lyes, lf(x, y)l and suppose that g has a sum overS'. Then f has a sum overS. We leave the proof of this and the following result to the reader.
Theorem 9.53 (Multiplication of unordered sums). Let f: S'-+ lll 1 and g: S"-+ lll 1 be given. Define S = S' x S" and h(x, y) = f(x) · g(y) for xeS', yeS". Suppose that f has a sum overS' and g has a sum overS". Then h has a sum over S and
I
(x,y)eS
h(x, Y> =
[I
xeS'
f(x>]·[
I
yeS"
g(y>]·
We now show how the theorems of this section may be used to establish rules for the multiplication of power series. Suppose the series
f(x) = a0 + a 1 x + · · · + anx" + · · ·, g(x) = b0
+ b1 x + · · · + bnx" + · · ·,
are convergent for lxl < R. Without considering questions of convergence, we multiply the two series by following the rules for the multiplication of polynomials. The result is
b0 f(x) = a0 b0 + a 1 b0 x + · · · + an box" + · · ·, b1 xf(x) bnx"f(x) = Adding, we obtain the series
a0 b0 + (a 0 b1 + a 1 b0 )x + · · · + (a 0 bn + a 1 bn-l + · · · + anb0 )x" + · · · .
9.7. Multiple Sequences and Series
259
In summation notation, the product becomes oo
oo
aooo
1=0
j=O
i=O j=O
L a1x 1 L b1xi = L L a1b1xi+i.
Setting i + j = n and collecting terms, we obtain for the right side
f (k.=Of ak.bn-k.) x".
11=0
We shall show that the above expansion actually represents f(x) · g(x). Definition. Let L~o a1(x - c)1 and the series
Lj;.o b1(x -
c)i be given power series. Then
Jo Cto a,.b,_,.) (x - c)"
(9.30)
is called the Cauchy product of the two given series. Theorem 9.54. Suppose that the series CIO
f(x) =
CIO
L c.,(x 11=0
c)",
g(x) =
L
d.,(x - c)",
11"'0
converge for lx - cl < R with R > 0. Then for lx- cl < R the product f(x) · g(x) is given by the Cauchy product of the two series. PROOF. Since the two given series are absolutely convergent for lx- cl < R, it follows that
f(x)
= L
c1(x - c)1,
g(x)
leNo
= L
d1(x - c)i.
jeNo
In Theorem 9.53, we takeS' = N 0 and S" = N 0 • Then the function his defined for (i, j) e N 0 x N 0 by hii = c1d1(x- c)i+i. Hence for each fixed x such that lx - cl < R, we have
L
f(x) · g(x) =
c1d1(x - c)l+i.
(l,j)e No x No
Now for each n e N 0 ,define S., = {(i, j) e N 0 x N 0 : i + j = n}. Then from the generalized associative law (Theorem 9.51) we find that the function u,(x)
=
L
c1d1(x - c)"
(l,j)eSn
has a sum over .N 0 • Therefore
f(x) · g(x)
= 11eLNo ( (l,j)eSn L c1d1) (x -
for lx - al < R. This formula is of the form (9.30).
c)"
D
Suppose that g is represented by a power series for lx- cl < R. Then it is
9. Infinite Sequences and Infinite Series
260
a consequence of Theorem 9.54 that if b is any real number, the expression [g(x)- b]" is given for lx- cl < R by the power series obtained by using the series for g and computing successive Cauchy products. In this way we get a convergent power series expansion for the composition of two function. If 00
f(x)
= L a"(x- b)" for lx- bi < R 0 with R 0 > 0, n=O
then we form the series f(g) =
a"(g - b)".
L
(9.31)
ne No
We now substitute successive Cauchy products for [g(x) - b]"into (9.31). The resulting series converges provided that lg(c)- bl < R 0 , and that lx- cl is sufficiently small. Thus f[g(x)] may be represented by a convergent power series in (x - c). Theorem 9.53 is useful for the extension to multinomial series of the results on double series. If the terms umn of a double series are of the form cmn(x - a)m(y - b)" where cmn• a, b, x, yare in IR 1 then we call 00
L
Cmn(x - a)m(y - b)"
m,n=O
a double power series. Triple, quadruple, and n-tuple power series are defined similarly. The proofs of the next two theorems follow directly from previous results on convergence.
Theorem 9.55. Suppose that the double power series L~.n=o cmnxmy" converges absolutely for x = x 1 , y = y 1 , and that x 1 =I= 0, y 1 =I= 0. DefineR= {(x,y): lxl ~ lx 1 I,IYI ~ IY 1 1}. Thentheseriesisabsolutelycon vergentonR,and the function with terms cmnxmy" has a sum uniformly on Rover N 0 x N 0 .
We may make the substitution x = x' - a andy = y' - bin Theorem 9.55 to obtain the analogous result for general double power series.
Theorem 9.56 (Term-by-term differentiation of double power series). Suppose that the double power series 00
f(x, y) =
L
Cmn(x - a)m(y - b)"
(9.32)
m,n=O
is absolutely convergent on S = {(x, y): lx- al < r, IY- bi < s}. Then f has partial derivatives of all orders on S. These derivatives may be obtained by differentiating the series (9.32) term by term. Each differentiated series is absolutely convergent on S. In particular,
1 am+nf I cmn = 1- ,-, --;nn x=a,y=b m.n.
ox oy
9.7. Multiple Sequences and Series
261
EXAMPLE 1. Find the terms of the double power series expansion to degree 3 where f(x, y) = exy, a = 1 b = 0.
Solution. We have f(1, 0) = 1. By computation, fx(1, 0) = 0, !,(1, 0) = 1, .fxx{1, 0) = 0, fxy(1, 0) = 1, Jyy(1, 0) = 1, fxxx(l, 0) = fxxy(1, 0) = 0, fxyy(1, 0) = 2, j,yy(1, 0) = 1. Therefore
exy
1 3!
1 2!
= 1 + y + (x- 1)y +- y 2 + (x- 1)y2 +- y 3 + ···.
D
EXAMPLE 2. Show that for all x, y, oo
kf:o
(x
+ y)k k!
( =
oo
xm)( x") oo
m'l;o m!
.f:o n! '
and hence that ex+y = exey.
Solution. According to Theorem 9.53, we have for all x, y, (
xmyn oo xm) ( oo y") . L:oo L:- = m,n=O L:m!n! n=O n!
m=O m!
Now we employ Theorem 9.50 to obtain (by setting n = p - m) 00
xmyn
00
xmyp-m
p
m,~o m!n! = pf:O m'l;o m!(p- m)!
f
=
p=O
(x
+ YY_
D
p!
PROBLEMS In each of Problems 1 through 6, find the terms up to degree 3 of the double series expansion (9.32) of f(x, y) as given. Take a = b = 0 in all cases. 1. f(x, y) = ex cos y 3. f(x, y) = e-x sec y 5. f(x, y) = cos(xy)
+ x2f 4. f(x, y) = e-lx log(1 + y) 6. f(x, y) = (1 + x + yfl/2 2. f(x, y) = (1
-X -
2y
1
7. State and prove the analogue of Theorem 9.47 for triple series. 8. Let s = N X N and suppose that u: s-+ ~ 1 is given by umn = 1/mn4 . Show that the double series L::;,n=l 1/mn4 is not convergent and hence that u does not have a sum overS. 9. Prove Theorem 9.49. 10. Write the details of the proof of Theorem 9.50 for W,: =
Lm+n=r umn·
11. Prove Theorem 9.51. Then letS'= Nand Sx = N x N for all x, and show that a theorem analogous to Theorem 9.50 can be derived for triple series.
262
9. Infinite Sequences and Infinite Series
12. Let 7l. denote the set of all integers: positive, negative, and zero. Apply Theorem 9.51 with S' = ~ 0 and Sx = 7l. for all x. Show how a theorem on convergence of "double series" of the form L~;o L.% -oo umn may be obtained. 13. Prove Theorem 9.52. 14. Suppose that in Theorem 9.52 we choose S' = ~. Sx = ~- We define f(m, n) = ( -1)m+n;(m 2 + n2 ). Are the hypotheses of the theorem satisfied? 15. Prove Theorem 9.53. 16. Use Theorem 9.53 to obtain a double power series expansion of ex cosy about (0, 0). 17. Prove Theorem 9.55. 18. State and prove a theorem on the term-by-term integration of double power series. 19. Prove Theorem 9.56. 20. Use power series expansions to show that cos(x
+ y) = cos x cosy - sin x sin y.
CHAPTER
10
Fourier Series
10.1. Expansions of Periodic Functions In the study of power series in Chapter 9, we saw that an analytic function f can be represented by a power series co
f(x) =
L cn(x n=O
a)"
(10.1)
for all values of x within the radius of convergence of the series. We recall that fhas derivatives of all orders and that the coefficients en in (10.1) are given by j;; x < 0}. Since we are interested in f only on J, properties of convergence of the series on I' are irrelevant. For example, we 0 on I'. However, one choice which is useful for many purposes may set f consists in defining f as an even function on I. Since bn = 0, n = 1, 2, ... , for even functions, the Fourier series will have cosine terms only. We call such a series a cosine series and, as the original function has domain J, the Fourier expansion is called a half-range series. A function f defined on J may be extended to I as an odd function. Then an= 0 for n = 0, 1, 2, ... , and the resulting series is called a sine series. We illustrate the process of obtaining cosine and sine series with two examples.
=
EXAMPLE
1. Given the function
f(x) = {0 for I 1 = {x: 0 ::>;; x < n/2}, 1 for I 2 = {x: n/2 ::>;; x ::>;; n}, find the cosine series for f Solution. We extend f as an even function, as shown in Figure 10.4. Then the function is standardized so that j(n/2) = j(3n/2) = j( -n/2) = !. Since j is even, we have bn = 0, n = 1, 2, ... Also,
2f" f(x) cos nx dx,
n = 0, 1, 2, ....
an = -
1t
0
y
0
0
•
•
311'
-2
-11'
11'
-2
I
2 0
0
0
•
• X
11'
2
11'
311'
T
Figure 10.4. Extension as an even function.
10.2. Sine Series and Cosine Series; Change of Interval
271
y
o----o
o-----o I
•
•
2
----~~--~---+----~--.---~--~~--~X
0
1r
31r
2
T
• _l
•
2
o-----o
-(
Figure 10.5. Extension as an odd function.
A simple calculation yields a 0 = 1,
an =
~I" n
cos nx dx = {
"12
~(
if n is even,
-1)k+l if n = 2k (2k + 1)n
+ 1, k =
0, 1, 2, ....
Therefore j(x),.., ~_~[cos x _cos 3x +cos 5x _ .. ·] 2n 1 3 5 '
D
EXAMPLE 2. Find the sine series for the function f for Example 1.
Solution. We extend f as an odd function as shown in Figure 10.5. The standardized function has j( -3n/2) = j(n/2) = · · · = j( -n) = j(n) = · · · = 0, and j( -n/2) = j(3nj2) = · · · = -t. Then an = 0 for n = 0, 1, 2, ... and
t.
2f" f(x) sin nx dx =-2f"
bn =Hence
l
1t
2
b
n
nn
=
1t
0
ifn is odd,
2
-[( -1)k- 1]
kn
sin nx dx.
Jt/2
ifn
= 2k and k = 1, 2, ....
10. Fourier Series
272
Therefore
) 2 [sin x 2 sin 2x sin 3x sin 5x ji-(X,..._ --- --+--+--
1
1t
-
2
2 sin 6x 6
3
+ -sin7-7x + "·
J
5
for
X E
I 1 U I 2•
D
Iff is a piecewise smooth function defined on an interval J = {x: c - 1t < x < c + 1t}, we may form the periodic extension off and compute the Fourier coefficients {an} and {bn} according to Formulas (10.4) and (10.5). It is clear that since the trigonometric functions have period 21t, these coefficients are also given by
an = -1 JO.
Solution. We shall compare f: x-+ xflj~ with g: x-+ 1/~. We have xfl xfl ~==
~
=
--;======
j(l-x)(1 +x)
~g(x).
yL.,..X
11.2. Convergence and Divergence oflmproper Integrals
293
J1+x for 0 ~ x ~ 1, it follows that 1/(x)l ~ g(x). By Example 1
Since xfl ~ we know that converges.
g 1/~ dx
is convergent, and so the original integral
0
We now take up the convergence of integrals in which the integrand is bounded but where the interval of integration is unbounded. Definitions. Let
f
be defined on I= {x: a~ x < +oo} and suppose that
J~f(x) dx exists for each c e /.Define
+oo Ic Ia f(x) dx = .~~co a f(x) dx whenever the limit on the right exists. In such cases, we say the integral converges; when the limit does not exist we say the integral diverges. Iff is definedonJ = {x: -oo < x < +oo}wemayconsiderexpressionsoftheform
J::
f(x)dx
which are determined in terms of two limits, one tending to +oo and the other to -oo. Let d be any point in J. Define
f+oo -co
f(x) dx = ••~~co
Id c,
f(x) dx + •2~~00
Ic2 d
f(x) dx
(11.9)
whenever both limits on the right exist. It is a simple matter to see that iff is integrable for every finite interval of J, then the values of the limits in (11.9) do not depend on the choice of the point d. To illustrate the convergence and divergence properties of integrals when the path of integration is infinite, we show that
I+oo a
x-p dx,
converges for p > 1 and diverges for p
I
c
a
x-p dx
~
a >0,
1. To see this, observe that (p =F 1)
1
= - - [c 1 - p - a 1-P]. 1- p
For p > 1, the right side tends to (1/(p- 1)) a 1 -p as c--+ +oo, while for p < 1 there is no limit. By the same argument the case p = 1 yields divergence. Similarly, the integral
foo lxl-p dx,
b 1 and diverges for p ~ 1. In analogy with the Comparison test for integrals over a finite path of integration, we state the following result.
294
11. Functions Defined by Integrals; Improper Integrals
Corollary. For continuous functions f and g defined on the interval I = { x: a ~ x < + 00} Theorems 11.4 and 11.5 are valid with respect to integrals J;oo f(x) dx and s;oo g(x) dx. EXAMPLE
3. Test for convergence:
f
+oo
1 JXJ/2 dx.
+x
1
Solution. For x
~
1, observe that
JX
JX_1_
1 + xJ/2 ~ 2x312 - 2x = g(x).
However
Jt
EXAMPLE
4. Test for convergence:
00
(1/2x) dx diverges and so the integral is divergent.
f-oo
D
+oo
Solution. Consider A and set u
xe-x2 dx.
=J: xe-x dx 2
= x 2, du = 2x dx. Then
rc2
1 1 1 A=2Jo e-"du=2-2e-c>, and A-+! as c-+
+oo. Since the integrand is an odd function we see that
f
o xe-x2 dx-+
_!
2
-d
as d-+
+oo, D
and the original integral is convergent.
It is clear that the convergence of integrals with unbounded integrands and over an infinite interval may be treated by combining Theorem 11.4 and the Corollary. The next example illustrates the method. EXAMPLE
5. Test for convergence:
i
+oo
0
e-x r:dx. yX
Solution. Because the integrand is unbounded near x = 0 we decompose the problem into two parts:
ioJX 1
e-x -dx
f
+oo
and
1
e-x r:dx. yX
11.3. The Derivative of Functions Defined by Improper Integrals
295
In the first integral we have e-x;Jx ~ 1/Jx, and the integral converges by the Comparison test. In the second integral we have e-x;Jx ~e-x, and once e-x dx is convergent. again the Comparison test yields the result since 0 Hence, the original integral converges.
Jioo
PROBLEMS
In each of Problems 1 through 12 test for convergence or divergence. 1.
f
3.
Jo Jl+?.
4.
5.
f-1Ji=7
6.
dx
+oo
+ 2)~·
(x
1
r+oo 1
7.
f
3
1
dx
r+oo
~
-1 -dx. ogx
8.
X
i
o
~
-.-dx. SID X
f 1 c-::z·
+oo (arctan x) 2 1+x
0
13. Show that
H"' x-
12. 1
o
i
2
dx.
dx
1
Jor+oo e-" sin x dx.
dx
Jo Ji+7. n/2
.
10.
11.
.
0
dx
dx
1J1=7 1
2.
v x- x 2
n
sin
o
XyX
X
r: dx.
(log xrP dx converges for p > 1 and diverges for p ~ 1.
14. Assume f is continuous on I = {x: 2 ~ x < + oo} and limx~ +oo x(log x) 2f(x) f(x) dx is convergent. and A # 0. Prove that
Ji"'
=A
11.3. The Derivative of Functions Defined by Improper Integrals; the Gamma Function We consider the integral
which we shall show is convergent for x > 0. Since the integrand is unbounded near zero when xis between 0 and 1, the integral may be split into two parts:
I+oo t"-le-1 dt = Il t"-le-1 dt + I+oo t"-le-1 dt =/1 + 12. In the first integral on the right, we use the inequality t"- 1 e- 1 ~ t"- 1 for
x > 0 and 0 < t
~ 1.
296
11. Functions Defined by Integrals; Improper Integrals
The integral f5 tx- 1 dt converges for x > 0 and so / 1 does also. As for 12 , an estimate for the integrand is obtained first by writing
and then estimating the function
f: t __. tx+1e-'. We find f'(t) = txe-'(x + 1 - t), and f has a maximum when t maximum value is f(x + 1) = (x + 1f+1e-<x+ 11. Therefore
~
(x
+ 1f+1e-<x+ 11
J
+oo
1
= (x
= x + 1. This
1 - dt t2
+ 1f+1e-<x+1J.
Hence / 2 is convergent for each fixed x > - 1.
Definition. For x > 0 the Gamma function, denoted r(x), is defined by the formula
The recursion formula r(x
+ 1) =
(11.10)
xr(x),
one of the most important properties of the Gamma function, is derived by means of an integration by parts. To see this, we write
J:
txe- 1 dt = [tx( -e-')] 0 + x
J:
tx- 1 e- 1 dt.
Now letting c--. +oo we obtain (11.10). It is easy to verify that r(1) = 1, and consequently that r(n + 1) = n! for positive integers n. Note that the Gamma function is a smooth extension to the positive real numbers of the factorial function which is defined only for the natural numbers. Leibniz's rule for differentiation of integrals was established for proper integrals. We require a more detailed study to establish similar formulas when the path of integration is infinite or when the integrand is unbounded. Let R = {(x, t): a ~ x ~ b, c ~ t < d} be a rectangle which does not contain its "upper'' side. Suppose that F: R--. ~ 1 is continuous and that
lim F(x, t) t-+d-
exists for each x e I= {x:
a~
x
~
b}. We denote the limit above by f(x).
297
11.3. The Derivative of Functions Defined by Improper Integrals
d
---r-J.
d-{j
c
L__
___
R
----- - - 1 - - - - - - - - - - i
a
0
b
Figure 11.2. F(x, t) -+ f(x) as {J
X
-+
0.
Defmition. The function F(x, t) tends to f(x) uniformly on I as t-+ d - if and only if for every 6 > 0 there is a {) > 0 such that IF(x, t) - f(x)l
0 there is a number T depending on 6 such that (11.11) holds for all t > T and all x on I. The number T depends on 6 but not on x. The next two theorems are the basis for Leibniz's rule for improper integrals.
Theorem 11.6. Suppose that F: R-+ R 1 is continuous on I for each t e J = {t: c ~ t < d}, and that F(x, t)-+ ,P(x) uniformly on I as t-+ d-. Then ,P is continuous on I. The same result holds if J is the half-infinite interval {t: c ~ t < +oo} and t-+ +oo.
PRooF. Let x 1 , x 2 be in I. We have ,P(x 1 ) - ,P(x2) = ,P(x 1 ) - F(xl> t)
+ F(x 1 , t)- F(x 2 , t) + F(x 2 , t)- ,P(x2)
and I,P(xd - ,P(x2)l
~
i,P(xd - F(xl> t)l
+ IF(xl, t)- F(x 2 , t)l + IF(x2, t)- ,P(x2)l.
11. Functions Defined by Integrals; Improper Integrals
298
Since F tends to
0 the first and third terms on the right can be made less than e/3 fort sufficiently close to d. If x 1 and x 2 are sufficiently close, the continuity of F assures us that the middle 0 term is less than e/3. Hence
Ycb f- 2e. 1
12.1. Functions of Bounded Variation
Therefore
v;,b f
307
n
L lf(x;) -
~
f(x;-dl >
v;,c f + ~b f -
e.
i=l
Since e is arbitrary, the result follows. EXAMPLE 1.
Show that
D
Suppose thatfis a nondecreasingfunction on I= {x: f(a) for x E I.
a~ x ~
b}.
v;,x f = f(x) -
Solution. Let x be in I and let L\ be a subdivision a= x 0 < x 1 < · · · < Xn = x. We have
0 be given. Then there is a subdivision A: a = x 0 < x 1 < · · · < xn = b such that
V..b f
;~ lf(x;)- f(x;-dl- 28.
12. The Riemann-Stieltjes Integral and Functions of Bounded Variation
310
Therefore, taking (12.2) into account,
V.t f- e =
0 be given. From the definition of integral, there is a
lit
lf'(OIIx;- X;-11-
r
lf'(x)l dxl
> 0 such that
~e
for every subdivision with mesh less than (>. Now we use the definition of bounded variation to assert that for the above e, there is a subdivision A0 : a= z0 < z 1 < · · · < Zm = b such that
Vabf Let A1 : a = xb
Va f- 2e. ;ft
< ···
~bf- 2B1 p
m
and
~i~ if(x;)- f(xj_t)l-
r
lf'(x)l dxl
is of bounded variation on I.
~
x ~ 1}. However, prove that
x 2 sin(1/x) for x # 0, {0 forx = 0,
315
12.1. Functions of Bounded Variation 8. Find those values of IX and
p for which the function
f :X-+ { x« sin(x- 11 ) for x * 0,
forx = 0,
0
is of bounded variation on I= {x: 0
~
x
~
1}.
9. Let f and g be functions of bounded variation on I = {x: a ~ x ~ b}. Show that
V.,b(f + g)
~
v..b f + v..b g, k a constant.
10. Let .1: a= x 0 < x 1 < · · · < x. = b be a subdivision of I= {x: a~ x ~ b} and suppose f is defined on I. Then Li'= 1 [(x;- X;-d 2 + (f(x;)- f(x;-df] 112 is the length of the inscribed polygonal arc off We define the length off on I, denoted L!f, by the formula L!f =sup
L• [(x;- X;-1)2 + (f(x;)- f(x;-1))2]112, i=l
where the supremum is taken over all possible subdivisions. (a) Show that for any function j, the inequalities
V.,b f + (b-
a)~
L!f ~ [(V.,b f)2
+ (b-
a)2]1f2
hold. Hence conclude that a function is of bounded variation if and only if it has finite length. (b) Show that if a< c < b then L!f = L~f + L~f [Hint: Follow the proof of Theorem 12.1.]
11. A function f defined on I = {x: a ~ x ~ b} is said to satisfy a uniform Holder condition with exponent IX if there is a number M such that lf(xd- j(x2)l
~
M ·lx1 - x2l«
for all x 1, x 2 e I.
(a) Iff satisfies a uniform Holder condition with IX = 1 show that f is of bounded variation. (b) Give an example of a function which satisfies a uniform Holder condition with 0 < IX < 1 and which is not of bounded variation. 12. Compute VO' f for f(x) = x•e-x on I = {x: 0
~
x ~ a}, where a> n > 0.
13. Complete the proof of Lemma 12.1. 14. Complete the proof of Theorem 12.2. 15. Show that iff is continuous and has a finite number of maxima and minima on an interval I= {x: a~ x ~ b}, thenfis of bounded variation on I. Conclude that every polynomial function is of bounded variation on every finite interval. 16. Suppose thatfis of bounded variation on I= {x: a~ x ~ b}. Iflf(x)l ~ c > Ofor all x e I where cis a constant, show that g(x) = 1/f(x) is of bounded variation on I.
316
12. The Riemann-Stieltjes Integral and Functions of Bounded Variation
12.2. The Riemann-Stieltjes Integral We introduce a generalization of the Riemann integral, one in which a function g. If g(x) = x then the generalized integral reduces to the Riemann integral. This new integral, called the Riemann-Stieltjes integral, has many applications not only in various branches of mathematics, but in physics and engineering as well. By choosing the function g appropriately we shall see that the Riemann-Stieltjes integral allows us to represent discrete as well as continuous processes in terms of integrals. This possibility yields applications to probability theory and statistics.
f is integrated with respect to a second function
Definitions.Letfandgbefunctionsfromi = {x: a~ x ~ b}into~ 1 .Suppose that there is a number A such that for each e > 0 there is a (j > 0 for which
~;~ f(C;)[g(x;)- g(x;-
1) ] -
AI< e
(12.7)
for every subdivision A of mesh size less than {J and for every sequence {0 with x;_ 1 ~(;~X;, i = 1, 2, ... , n. Then we say thatfis integrable with respect tog on I. We also say that the integral exists in the Riemann-Stieltjes sense. The number A is called the R-S integral off with respect tog and we write A
=
r r f
dg
=
f(x) dg(x).
As in the case of a Riemann integral, it is a simple matter to show that when the number A exists it is unique. Furthermore, when g(x) = x, the sum in (12. 7) is a Riemann sum and the R -S integral reduces to the Riemann integral as described in Chapter 5. It is important to observe that the R-S integral may exist when g is not continuous. For example, with I = {x: 0 ~ x ~ 1} let f(x) 1 on I,
=
g(x) = {0 for 1 for
?~
x
0 be given. We wish to show that
Iit f(C;)[g(x;)- g(X;-1)]-
f(x)g'(x)
dxl < e
(12.9)
provided that the mesh of the subdivision is sufficiently small. We apply the Mean-value theorem to the Riemann-Stieltjes sum in (12.9) getting n
n
i=1
i=1
L f(C;)[g(x;)- g(x;-1)] = L f(C;)g'('f;)(x;- X;-1)
(12.10)
where x 1_ 1 ~ 'I; ~ x 1• The sum on the right would be a Riemann sum if 'I; where equal to C1• We show that for subdivisions with sufficiently small mesh this sum is close to a Riemann sum. Let M denote the maximum of lf(x)l on I. Since g' is continuous on I, it is uniformly continuous there. Hence there is a {J > 0 such that for 1'1 - 'l;l < {J it follows that (12.11) From the definition of Riemann integral there is a subdivision with mesh so small (and less than {J) that
I,~ f(C;)g'(C;)(X;- X;-1)-
r
f(x)g'(x)
dxl 0 be given. From the definition ofR-S integral, there is a{> > 0 such that (12.16) for all subdivisions A: a = x0 < x 1 < · · · < xn = b with mesh less than {> and any {(;}in which x;_ 1 ~ '; ~ X;. Since xis (uniformly) continuous on J, there is a {> 1 such that lx(u')- x(u")l < {> whenever lu'- u"l < {> 1 • Consider the subdivision A1 : c = u0 < u 1 < · · · < un = d of J with mesh less than {> 1 . Let '1; be such that u;_ 1 ~ '1; ~ u; for i = 1, 2, ... , n. Since x(u) is increasing on J, denote x(u;) = x; and x(l'f;) = ';· Then the subdivision A of I has mesh less than {> and X;_ 1 ~ '; ~ X; for i = 1, 2, ... , n. Therefore n
n
1=1
i=1
L F(l'f;)[G(u;)- G(u;-d] = L f(,;)[g(x;)- g(x;-d].
320
12. The Riemann-Stieltjes Integral and Functions of Bounded Variation
Taking into account the definition of the R-S integral, (12.7), we see that the integrals in (12.14) are equal. If x(u) is decreasing on J, let x(u;) = x,_;, x('l;) = 1 -;. Then
,
,
~1
~1
,,+
L F('l;)[G(u;)- G(u;-1)] = L /(,,+ 1-;)[g(x,_;)- g(x,+ 1-;)].
Setting k = n + 1 - i, i
= 1, 2, ... , n, we find the sum on the right is ,
- k=1 L f(,k)[g(xk)- g(xk-1)].
Then (12.16) shows that the integrals in (12.15) are equal.
D
The next result is a generalization of the customary integration by parts formula studied in calculus. Formula (12.17) below is most useful for the actual computation ofR -S integrals. It also shows that if either of the integrals J!f dg or g df exists then the other one does.
J!
Theorem 12.12 (Integration by parts). I/ J!f dg exists, then so does J!g df and
r
g df = g(b)f(b) - g(a)f(a) -
r
f dg.
(12.17)
8 > 0 be given. From the definition of R -S integral, there is a > 0 such that
PRooF. Let {l
.t '
f(W[g(x;)- g(xi-1)]-
fb f
dg'
0 there is an integer N such that d(pm, p,) < 8 whenever m, n > N. We also recall Theorem 6.22 in Chapter 6 which states that every convergent sequence in a metric space is a Cauchy sequence. Since not every Cauchy sequence in a metric space is convergent, we introduce the following class of metric spaces. Definition. A metric space S is said to be complete if and only if every Cauchy sequence inS converges to a point inS. We observe that a compact metric space is always complete (Theorem 6.23). Although the space IR 1 is not compact it is complete, as was established in Theorem 3.14. With the aid of this result we now show that all the Euclidean spaces are complete. Theorem 13.1. For every positive integer N, the space IRN is complete. We already showed that IR 1 is complete. Let N ~ 2. Let x 1 , x 2 , ... , x,, ... be a Cauchy sequence in IRN. With the notation x, = (x~, x;, ... , x:), PROOF.
329
330
13. Contraction Mappings, Newton's Method, and Differential Equations
the distance between the points xm and x,. is d(xm, x,.) =
[
N
.L (x~ -
]1/2 x!) 2 .
•=1
Fix j and observe that for each j = 1, 2, ... , N, N
L (x~- x!)2 ;;;l!: (x!- x~) 2 . i=1 Hence for each fixed j with 1 ~ j ~ N the sequence {x~} is a Cauchy sequence in IR 1 and thus converges to a limit which is denoted x6. We conclude that x,. -+ x 0 where x 0 = (x~, x~, ... , xg}. The space IRN is complete. D Definition. Let f be a mapping from a metric space S into S. A point x 0 in S is called a fixed point of the mapping f if f(x 0) = x 0 • We now show that certain classes of mappings from a complete metric space into itself always have a fixed point. Theorem 13.2 (A fixed point theorem). Suppose that f is a mapping on a complete metric space S into S such that d(f(x), f(y))
~
k d(x, y)
(13.1)
for all x, y in S with 0 < k < 1. Then the mapping has a unique fixed point. That is, there is a unique point x 0 e S such that f(x 0) = x 0 • Moreover, if x 1 is any element of S, and the sequence {x,.} is defined by setting x,.+ 1 = f(x,.), n = 1, 2, ... , then x,. converges to x 0 • PROOF. Start with any x 1 in S and set x,.+ 1 = f(x,.), n = 1, 2, .... Then d(x 2, x 3) = d(f(xd, f(x 2)) ~ kd(x 1, x 2). Continuing,
d(x 3, x 4) = d(f(x 2), f(x 3)) ~ kd(x 2, x 3) ~ k 2d(x 1, x2). In general, d(x,., x,.+d ~ k"- 1d(x 1, x 2 )
for
n = 1, 2,....
(13.2)
Let m, n be any positive integers with m > n. Then by the triangle inequality in a metric space, d(x,., Xm) ~ d(x,., Xn+1)
+ d(xn+l• Xn+2) + ··· + d(Xm-1• Xm).
Applying inequality (13.2) to each term on the right, we obtain
+ k" + ... + km-2] d(x1, x2)k"- 1[1 + k + ··· + km-n- 1].
d(x,., Xm) ~ d(x1, x2)[kn-1 =
Since L.i=o ki = 1/(1 - k), we find d(x,., xm) < d(x 1, x 2)k"-11 ~ k.
331
13.1. A Fixed Point Theorem and Newton's Method
Now let 6 > 0 be given. Since 0 < k < 1, the right side tends to 0 as n -+ oo. Therefore there is an N such that the right side is less than 6 for all n > N. Thus {x,.} is a Cauchy sequence. Let x 0 be the limit of {x,.}. By (13.1), we know that f is continuous on S and so f(x,) -+ f(x 0 ) as n -+ oo. Since x,.+ 1 -+ x 0 and x,.+ 1 = f(x,), it follows that f(x 0 ) = x 0 • To establish the uniqueness ofthe point x 0 , suppose there is a point x' such that x' = f(x'). Using (13.1), we find that d(x 0 , x') = d[f(x 0 ), f(x')]
~
kd(x 0 , x').
Since 0 < k < 1, it is clear that d(x 0 , x') = 0, and so x 0 = x'.
D
A mapping f which satisfies (13.1) with 0 < k < 1 is called a contraction mapping. Corollary.Let I= {x: a ~ x ~ b} be an interval in IR. 1 and suppose thatf: I-+ I is a differentiable function at each interior point of I with lf'(x)l ~ k, 0 < k < 1 for each x. Then there is a unique point x 0 of I such that f(x 0 ) = x 0 • We leave the proof to the reader. EXAMPLE 1. Let/ = { x: a ~ x ~ b} be an interval and suppose that f is differentiable on I with f(a) and -f(b) contained in the interval J = {x: 0 ~ x ~ b- a}. Furthermore, suppose that there are numbers k and k' such that -1 < k' ~ f'(x) ~ k < 0 for all x e I. Show that there is an x 0 e I such that f(x 0 ) = 0 and prove that for any x 1 e I, the iteration x,.+ 1 = f(x,), n = 1, 2, ... converges to x 0 •
Solution. Set g(x) = f(x) + x. Then, by hypothesis, g(a) e I, g(b) e I, and, since g'(x) = 1 + f'(x), it follows that g is an increasing function on I. Hence g maps I into I. Also, g(x) - g(y) = x - y + f(x) - f(y) and, applying the Mean-
value Theorem, we find lg(x) - g(y)l = l(x - y)(1 with a < C< b. Thus lg(x)- g(y)l ~ (1
+ f'(O)I
+ k)lx- yl,
and from Theorem 13.2 it follows that there is an x 0 such that g(x 0 ) Therefore x 0 + f(x 0 ) = x 0 and f(x 0 ) = 0.
= x0 • D
Remarks. The above example shows that for any differentiable function J, if f(a) and f(b) are of opposite sign and if the derivative off is small (but bounded away from zero) then the zero off which must exist between a and b may always be found by a simple iteration method. There are many iteration methods for finding the zeros of functions on an interval. We now describe a method of wide applicability, especially when the first derivative of the function is large in the neighborhood of the zero.
332
13. Contraction Mappings, Newton's Method, and Differential Equations y
0
a
b y =[(x)
Y =[(Xn) + f'(xn) (x -x.)
Figure 13.1. Newton's method.
Newton's method
Let f be a twice differentiable function on I = { x: a ::::; x ::::; b} with f(a) > 0 and f(b) < 0. Then there is an x 0 E I such that f(x 0 ) = 0. Newton's method consists of an iteration process which converges to x 0 , one of the zeros off in I. We assume that there is a positive number M such that lf'(x)l ~ 1/M and lf"(x)l ::::; 2M for all x E I. Let x 1 be any point in I. We define a sequence {xn} by the formula, known as Newton's method, f(xn) Xn+l = Xn - J'(xn)'
n
= 1, 2, ....
(13.3)
The determination of xn+l from xn is shown geometrically in Figure 13.1. The tangent line to the graph of y = f(x) is constructed at the point (xn, f(xn)). The point where this line intersects the x axis is xn+l· To show that the sequence {xn} converges to x 0 , first observe that (13.4)
lxn+l- xnl::::; MIJ(xn)l. Next, apply Taylor's theorem with remainder to fat xn+l, getting f(xn+d = f(xn)
+ f'(xn)(xn+l
- Xn)
+ !J"(()(xn+l -
Xn) 2 •
Because of (13.3) the first two terms on the right cancel, and so IJ(xn+l)l::::; Mlxn+l- Xnl 2 •
(13.5)
By means of(13.4) and (13.5), it is not difficult to show by induction that {xn} is a convergent sequence provided lf(xdl and M are less than 1. Hence it
333
13.1. A Fixed Point Theorem and Newton's Method
follows from (13.5) that f(xn)-+ 0 as n-+ oo. In terms of the Fixed point theorem, we define the function 1
F(x) = x - f'(x/(x); then Newton's method consists of forming xn+t = F(xn), n = 1, 2, ... , and concluding from the convergence of the sequence that there is a fixed point x 0 such that F(x 0 ) = x 0 • The existence of such a point in general does not follow from Theorem 13.2 since the range ofF is not necessarily contained in its domain and so we may not have a mapping of a complete metric space I into itself. 2. Find a positive root of the equation 2x 4 + 2x 3 5 = 0 with an accuracy of three decimal places.
EXAMPLE
Solution. We construct a table of values of f(x) = 2x4 and obtain f;x)
+ 2x 3 -
-
3x 2 3x 2
-
-
5x 5x - 5
~-~ ~-~ ~2~ I·
Thus there is a positive root x 0 between x = 1 and x = 2. We choose x 1 = 1.6 as a first approximation and apply Newton's method. We calculate f'(x) = 8x 3 + 6x 2 - 6x - 5. Then, according to Formula (13.3) we have
x2
=
0.6193 f(l.6) 1.6- f'(l. 6) = 1.6- 33 .528 = 1.5815.
The next approximation yields x3 = x2
-
f(x 2 ) 0.114 f'(x 2 ) = 1.5815 - 32.1623 = 1.5780.
y
Y = f(x 1 )
+ f'(x 1 )(x- x,)
Figure 13.2. Newton's method fails because x 1 is not close to x 0 •
334
13. Contraction Mappings, Newton's Method, and Differential Equations
Continuing, we find
x4 = x 3
f(x3)
-
f'(x 3 ) = 1.5780 + 0.0031 = 1.5811.
0
The root, accurate to three decimal places is 1.581.
Newton's method may fail if the first approximation x 1 is not sufficiently close to the root x 0 or if the slope off at x 1 is not sufficiently large. Figure 13.2 illustrates how the second approximation, x 2 , may be further from the root than the first approximation, x 1 . PROBLEMS
1. Find the positive root of 2x 4 places.
3x2
-
5
-
= 0 with an accuracy of three decimal
2. Find the positive root of 3x 3 decimal places.
+ 6x 2
-
3. Find the negative root of x 4 decimal places.
2x 3
3x 2
4. Find the root of x - cos x places. 5. Find all the roots of 3x 3 places.
-
-
= 0, 0 ~ x
+ x2
6. Find the positive root of 3x 3 decimal places.
11x
-
7x- 14
~
-
= 0 with an accuracy of three
2x - 4
= 0 with an accuracy of two
n/2 with an accuracy of two decimal
+ 6 = 0 with an
accuracy of two decimal
+ 16x 2 - 8x- 16 = 0 with an accuracy of four
7. Given f: x -+ (1/4)(1 - x - (1/10)x 5 ) defined on I = {x: 0 ~ x ~ 1}. By applying the Fixed point theorem to g(x) = f(x) + x, show that f has a zero in I. Set up an iteration process and find the first three terms. 8. Consider the function f: x -+ jl+? defined on I = {x: 0 ~ x < oo}. Show that lf(x)- f(y)l < lx- yl for all x of. y and that f does not have a fixed point. Conclude that Theorem 13.2 is false if k = 1. 9. Prove the Corollary to Theorem 13.2. 10. Let f: (x, y)-+ (x', y') be a function from IR 2 to IR 2 defined by
x' =ix +h-2, y'
=!x-h+ 3.
Use Theorem 13.2 to show that f has a fixed point. 11. Let f: (x, y)-+ (x', y') be a function from IR 2 to IR 2 defined by
x' = 1sin x- 1cosy+ 2, y' =
i cos x + t sin y -
Use Theorem 13.2 to show that f has a fixed point.
1.
13.2. Application of the Fixed Point Theorem to Differential Equations 12. Let
j2
f be a mapping of a complete metric space
335
S into S, and suppose that
= f of satisfies the conditions of Theorem 13.2. Show that fhas a unique fixed
point. 13. Use Newton's method to find to three decimal places the root of e-x- 5x = 0 on the interval I= {x: 0 ~ x ~ 1}. 14. Complete the details of the induction in the proof of convergence of Newton's
method.
15. Suppose that an iteration, similar to Newton's method, is used according to the formula f(x.) n = 1, 2, .... Xn+l = x.- f'(xd' State and prove a theorem which establishes convergence of this method.
13.2. Application of the Fixed Point Theorem to Differential Equations Let D be an open set in IR 2 and suppose that f: D--+ IR 1 is continuous. We now investigate the problem of determining solutions in D of the first order differential equation
dy dx
= f(x, y)
(13.6)
by means of the fixed point theorem of Section 13.1.
Definitions. Let P(x 0 , y 0 } be a point of D. A function y = qJ(x) is a solution of the initial value problem of (13.6) if and only if qJ'(x) = f(x, qJ(x)) for x in some interval I= {x: x 0 - h < x < x 0 + h} and if qJ satisfies the initial condition Yo= qJ(Xo).
We shall show that when f satisfies certain smoothness conditions, a solution to the initial value problem exists and is unique. Since the method employs the Fixed point theorem of Section 1, the solution may always be found by an iteration technique.
Definitions. LetS be a metric space and suppose thatf: S--+ IR 1 and g: S--+ IR 1 are bounded continuous functions on S. We define the distance between f and gby (13.7) d(f, g) = sup lf(x) - g(x)\. xeS
It is easy to verify that the collection of all bounded continuous functions from S to IR 1 forms a metric space with the distance function given by (13.7). We denote this space by C(S).
336
13. Contraction Mappings, Newton's Method, and Differential Equations
Theorem 13.3. Let S be a metric space and C(S) the metric space of bounded continuous functions on S. Let {f,} be a sequence of elements of C(S). Then f,--+ fin C(S) as n--+ oo if and only iff,--+ f uniformly on S as n--+ oo. Suppose f, converges to f uniformly on S. Thus, suppose that for every e > 0 there is an integer n0 such that PRooF.
lfn(x) - f(x)l < e for all and the number n0 is independent of x sup lfn(x) - f(x)l
> n0 ,
S. Consequently,
E
~
n
e for
n > n0 ,
xeS
and we conclude that d(f,, f)--+ 0 as n--+ oo. That is, fn converges to fin C(S). D The converse is obtained by reversing the steps in the argument. We leave to the reader the proof of the next two theorems.
Theorem 13.4. If S is any metric space, the space C(S) is complete. Theorem 13.5. Let S be a complete metric space and A a closed subset of S. Then A, considered as a metric space, is complete. Remarks. LetS be a metric space and suppose thatf: S--+ IRN and g: S--+ IRN are bounded continuous functions on S. That is, f and g have the form f = (f, f2, ... ,JN), g = (gl, g2, ... , gN) where /;, gi, i = 1, 2, ... , N, are bounded continuous functions from S into IR 1 • If we define lf(x)- g(xW = l:f= 1 1/;(x) - gi(xW, then the distance formula (13. 7) determines a metric space denoted CN (S). It is now a straightforward matter to establish the analog of Theorem 13.3 for the space CN(S). The next lemma enables us to reduce the problem of solving a differential equation to that of solving an integral equation. This reduction is used extensively in problems involving ordinary and partial differential equations.
Lemma 13.1. Suppose that f: D--+ IR 1 is continuous, that qJ is defined and continuous on I = {x: x 0 - h < x < x 0 + h} to IR 1 , and that (x 0 , y 0 ) E D with qJ(x 0 ) =Yo· Then a necessary and sufficient condition that qJ be a solution of dqJ dx = f[x, ({J(x)]
(13.8)
on I is that qJ satisfy the integral equation
qJ(x) = Yo +
I:
f[t, ({J(t)] dt
for
x E I.
(13.9)
We integrate (13.8) between x 0 and x obtaining (13.9). Then the D Fundamental theorem of calculus establishes the required equivalence.
PROOF.
337
13.2. Application of the Fixed Point Theorem to Differential Equations
Definitions. Let F be defined on a setS in IR 1 with values in IR 1• We say that F satisfies a Lipschitz condition on S if and only if there is a constant M such that (13.10) for all values of x 1 , x 2 inS. The smallest number M for which (13.10) holds is called the Lipschitz constant.
Iff has a bounded derivative on an interval I, then it satisfies a Lipschitz condition there. To see this, observe that by the Mean-value theorem there is a value' such that f(xd- f(x 2) = f'(0(x 1 - x 2) for any x 1 , x 2 E I. If 1!'(01 ~ M for all' E I, then (13.10) clearly holds for the function f On the other hand, a function may satisfy a Lipschitz condition and not be differentiable at certain points of I. A function whose graph consists of several connected straight line segments illustrates this property.
Theorem 13.6. Suppose that f: R-+ IR 1 is continuous where R is the rectangle R = {(x, y): lx- x 0 1< h, IY- Yo I~ k}, that lf(x, y)l ~ M 0 for (x, y) E R, and that f satisfies a Lipschitz condition with respect to y,
lf(x, Yd- f(x, Y2)l
~ MtiYt - Y2l
whenever (x, yd and (x, y2 ) are in R. Then M0 h ~ k
and
if M 1 h < 1,
there is a unique continuously differentiable function qJ defined on I = {x: x 0 - h < x < x 0 + h} for which dqJ I({J(x)- Yo I ~ k and dx = f[x, qJ(x)]. PROOF. According to Lemma 13.1 it suffices to find a solution of the integral equation
qJ(x)
= Yo +
Ix f[t, ({J(t)] dt
(13.11)
Xo
with I qJ(x) - Yo I ~ k for x E I. Let C(I) be the metric space of bounded continuous functions on I, and let E be the subset of C(I) for which I qJ(x) - Yo I ~ k.lf { ({Jn} is a sequence of functions in E which converges in C(I) to a function qJ0 , then the inequality 1(/Jo- Yol ~ 1(/Jo- (/Jnl
+ I({Jn- Yol
shows that ({Jo must also belong to E. Thus E is a closed subset of C(I) and, by Theorem 13.5, a complete metric space. We define a mapping TonE by the formula
T(qJ) = t/1
where
t/J(x) =Yo
+
Ix f[t, ({J(t)] dt, Xo
qJ
E
E.
338
13. Contraction Mappings, Newton's Method, and Differential Equations
We have lt/J(x)- Yo I
=I I:
f[t, q>(t)] dtl
~M
0
Jx- x 0 1 < M 0 h
~ k.
Hence t/1 is in E and the mapping T takes the metric space E into itself. Now we show that Tis a contraction map. Let q> 1, q> 2 be in E and denote t/1 1 = T(q> 1), t/1 2 = T(q> 2 ). Then lt/12(x)- t/11(x)l
=If~ {f[t, IP2(t)]- f[t, q> 1(t)]} dtl
~II: M111P2(t)-tp1(t)Jdtl tel
Therefore d(t/1 2 , t/1 1) ~ M 1 hd(q> 2 , tpt}. Since M 1 h < 1 by hypothesis, Theorem 13.2 shows that the mapping T has a unique fixed point. That is, Equation D (13.11) has a unique solution and the result is established. Remarks. Iff is defined in an open set Din IR 2 and M 0 , M 1 of Theorem 13.6 hold for all of D, then the solution q>, valid in a rectangle, can be extended throughout D. We observe that in the proof of Theorem 13.6, the size of h depends only on M 0 , M 1 and k and not on q>. Once a solution q> is found in a rectangle centered at (x 0 , y0 ), we may take any other point on this solution curve, say (x 1 , yt) and solve the initial value problem in a new rectangle centered at (x 1 , yt) (see Figure 13.3).11 is not difficult to show that in general this new solution will extend beyond the original one. By the uniqueness result, the solutions must coincide in the overlapping portions of the rectangles. Proceeding step by step we obtain a solution of the initial value problem throughout D.
~--------h--------~
Figure 13.3. Extending a solution.
13.2. Application of the Fixed Point Theorem to Differential Equations
339
Theorem 13.6 has an immediate extension to solutions of the system of equations dy; i = 1, 2, ... , N. dx = J;(x, Yt• Y2•. · ·, YN), We consider mappings from I into RN and the corresponding space of functions CN(J) as described in the remarks following Theorem 13.5. We obtain an existence and uniqueness theorem for the initial value problem provided the J; are continuous and satisfy a Lipschitz condition with respect to each of the variables Yt• Y2• ... , YN· The uniqueness conclusion of Theorem 13.6 fails if we drop the hypothesis of a Lipschitz condition and merely assume that f is continuous. To see this, consider the equation dy = yl/3 (13.12) dx
=
in any open set containing (0, 0). Then it is clear that y = cp(x) 0 is a solution to (13.12) which satisfies the initial condition cp(O) = 0. A second solution is given by y = cp(x) = (2x/3) 3' 2 • It is not difficult to see that the function f(x, y) = y 1!3 does not satisfy a Lipschitz condition in any open set which contains the point (0, 0). PROBLEMS
1. Let S be a metric space. Show that the space C(S) is complete (Theorem 13.4). 2. Let S be a complete metric space and A a closed subset of S. Show that A, considered as a metric space, is complete (Theorem 13.5).
3. Prove Theorem 13.3 with the metric space CN(S) in place of the space C(S). 4. Consider the integral equation
cp(x)
= g(x) + A.
r r
K(x, y)cp(y) dy,
where A. is a constant, g is continuous on I= {x: a :E;; x :E;; b}, and K is continuous on the squareS= {(x, y): a :E;; x :E;; b, a :E;; y :E;; b}. Define the mapping 1/1 = Tcp by 1/1 = Tcp
= g(x) + A.
K(x, y)cp(y) dy.
Use the fixed point theorem to show that for sufficiently small values of A. there is a unique solution to the integral equation.
5. Given the differential equation dyfdx = x 2 + y 2 and the initial condition cp(O) = 1, use the method of reduction to an integral equation and successive approximation to find the first six terms in the Taylor expansion solution y = cp(x). 6. In the proof of Theorem 13.6, write a complete proof of the statement that E is a closed subset of C(l).
340
13. Contraction Mappings, Newton's Method, and Differential Equations
7. Given the linear differential equation dyjdx = A(x)y + B(x), show that if A(x) and B(x) are bounded and integrable on I= {x: a~ x ~ b}, then the fixed point theorem yields a solution to the initial value problem on I. 8. Given the second order linear equation d 2y dx 2
dy
+ A(x) dx + B(x)y + C(x) = 0.
Let y 1 = y and y 2 = dyjdx. Then the second order equation reduces to the pair of first order equations: dyz dx
= -A(x)y 2
If A, B, and Care continuous on I= {x: to Theorem 13.6 holds.
a~
x
C(x).
-
B(x)y 1
~
b}, show that a theorem similar
-
9. Consider the system of differential equations dy; dx
= J;(x, Yt, Yz, · · ·, YN),
i
= 1, 2, ... , N.
(13.13)
State and prove the analog of Lemma 13.1 for such a system. 10. State and prove the analog of Theorem 13.6 for systems of the form (13.13).
CHAPTER
14
Implicit Function Theorems and Lagrange Multipliers
14.1. The Implicit Function Theorem for a Single Equation Suppose we are given a relation in ~ 2 of the form F(x, y) = 0.
(14.1)
Then to each value of x there may correspond one or more values of y which satisfy (14.1)-or there may be no values of y which do so. If I= {x: x 0 - h < x < x 0 + h} is an interval such that for each x e I there is exactly one value of y satisfying (14.1), then we say that F(x, y) = 0 defines y as a function of x implicitly on I. Denoting this function by J, we have F[x, f(x)] = 0 for x on I. An Implicit function theorem is one which determines conditions under which a relation such as (14.1) defines y as a function of x or x as a function of y. The solution is a local one in the sense that the size of the interval I may be much smaller than the domain of the relation F. Figure 14.1 shows the graph of a relation such as (14.1). We see that F defines y as a function of x in a region about P, but not beyond the point Q. Furthermore, the relation does not yield y as a function of x in any region containing the point Q in its interior. The simplest example of an Implicit function theorem states that if F is smooth and if Pis a point at which F, 2 (that is, oFfoy) does not vanish, then it is possible to express y as a function of x in a region containing this point. More precisely we have the following result. Theorem 14.1. Suppose that F, F, 1 and F, 2 are continuous on an open set A in ~ 2 containing the point P(x 0 , y 0 ), and suppose that F(xo, Yo) = 0,
341
342
14. Implicit Function Theorems and Lagrange Multipliers y
Q
F(x,y) = 0 --~-----------------------------x 0
Figure 14.1 (a) Then there are positive numbers h and k which determine a rectangle R
contained in A (see Figure 14.2) given by
R = {(x, y): lx- x0 1< h, IY- Yol < k}, such that for each x in I= {x: lx- x0 1< h} there is a unique number yin J = {y: IY- Yo I < k} which satisfies the equation F(x, y) = 0. The totality of the points (x, y) forms a function f whose domain contains I and whose range is in J. (b) The function f and its derivative f' are continuous on I. We shall give two proofs of Part (a), one which uses the elementary propery
Yo +k
- - - - - - - - 1- - - - ----F(x,y)>O
I I Yo F(x,y) =0
R Yo-k
- - - - + - - -.........- - - - - !1----F(x,y) 0; otherwise we replace F by - F and repeat the argument. Since F, 2 is continuous there is a (sufficiently small) squareS= {(x, y): Jx- x 0 J ~ k, Jy- Yol ~ k} which is contained in A and on which F, 2 is positive. For each fixed value of x such that Jx- x 0 J < k we see that F(x, y), considered as a function of y, is an increasing function. Since F(x 0 , y0 ) = 0, it is clear that
F(x 0 , Yo + k) > 0
and
F(x 0 , Yo - k) < 0.
Because F is continuous on S, there is a (sufficiently small) number h such that F(x, Yo+ k) > 0 on I= {x: Jx- x 0 J < h} and F(x, y 0 - k) < 0 on I. We fix a value of x in I and examine solutions of F(x, y) = 0 in the rectangle R (see Figure 14.2). Since F(x, Yo- k) is negative and F(x, y 0 + k) is positive, there is a value yin R such that F(x, Y) = 0. Also, because F, 2 > 0, there is precisely one such value. The correspondence x -+ y is the function we seek, and we D denote it by f (b) To show that f is continuous at x 0 let 8 > 0 be given and suppose that is smaller than k. Then we may construct a square with side 28 and center at (x 0 , y 0 ) as in the proof of Part (a). There is a value h' < h such that f is a function on I'= {x: Jx- x 0 J < h'}. Therefore
s.
8
Jf(x)- f(x 0 )l
'
c- -=':...::..-,--__:__'----"-:-
1/J(x, y) =
(
1
F,2 Xo, Yo
) [F(x, y) - F, 1(x 0 , y 0 )(x - Xo)
- ~2(xo, YoHY- Yo)]. Then the mapping TxY can be written
TxY = Yo - c(x - Xo) - 1/J(x, y). Since F(x 0 , y 0 )
= 0, we see that 1/1.1 (xo, Yo) = 0,
Because 1/1, 1 and 1/1. 2 are continuous we can take k so small that ll/l.2(x, y)l :;;;;
t,
for (x, y) in the square S = {(x, y): lx- x 0 1:;;;; k, IY- Yo I:;;;; k}. We now expand 1/J(x, y) in a Taylor series in S about the point (x 0 , y 0 ) getting
1/J(x, y)
= 1/1.1(~,
17)(x- Xo) + 1/1.2(~, '1HY- Yo),
(~,
17) E S.
Hence for h :;;;; k, we have the estimate in the rectangle R: 11/J(x, y)l :;;;;
!h + tk.
Next we show that if we reduce h sufficiently, the mapping T,., takes the interval (space) J into J. We have
IT,.,y- Yo I :;;;; lc(x- Xo)l
:;;;; lclh + !h
+
+ 11/J(x, y)l
!k =
(! + lcl)h +
tk.
We choose h so small that(!+ lcl)h :;;;; k. Then T,.,y maps J into J for each x in I= {x: lx- x 0 1:;;;; h}. The mapping T,., is a contraction map; in fact, by the Mean-value theorem
14.1. The Implicit Function Theorem for a Single Equation
345
We apply Theorem 13.2 and for each fixed x in I, there is a unique yin J such D that F(x, y) = 0. That is, y is a function of x for (x, y) E R. The Implicit function theorem has a number of generalizations and applications. IfF is a function from IRN+ 1 to IR\ we may consider whether or not the relation F(x 1 , x 2 , ••• , xN, y) = 0 defines y as a function from IRN into IR 1 • That is, we state conditions which show that y = f(x 1 , x 2 , ••• , xN). The proof of the following theorem is a straightforward extension of the proof of Theorem 14.1 and we leave the details to the reader. Theorem 14.2. Suppose that F, F, 1 , F, 2 , ... , F,N+l are continuous on an open set A in IRN+ 1 containing the point P(x~, x~, ... , x~, y 0 ). We use the notation x = (x 1 , x 2 , ••• , xN), x 0 = (x~, x~, ... , x~) and suppose that
(a) Then there are positive numbers hand k which determine a cell R contained in A given by
i = 1, 2, ... , N, IY- y 0 l < k},
R = {(x, y): lxi- x?l < h,
such that for each x in theN-dimensional hypercube IN={x:lxi-x?l 0 for all x, y. Hence for each fixed x, the function F is an increasing function of y. Furthermore, from (14.5) it follows that F(x, y)-+ -oo as y-+ -oo and F(x, y)-+ +oo as y-+ +oo. Since F is continuous, for each fixed x there is exactly one value of y such that F(x, y) = 0. Applying Theorem 14.1, we conclude that there is a function f on lll 1 which is continuous and differentiable such that F[x, f(x)] = 0 for all x. 0 EXAMPLE
2. Given the relation F(x, y) = x 3
+ y3 -
6xy = 0,
(14.6)
find the values of x for which the relation defines y as a function of x (on some interval) and find the values of y for which the relation defines x as a function of y (on some interval).
Solution. The graph ofthe relation is shown in Figure 14.3. We see that F, 1 = 3x 2
-
6y,
F, 2
= 3y 2 - 6x,
and both partial derivatives vanish at (0, 0). We also observe that F, 2 = 0 when x = ty 2 , and substituting this value into the relation (14.6) we get x = 2.y4, y = 2~. The curve has a vertical tangent at this point, denoted P in Figure 14.3. Hence y is expressible as a function of x in a neighborhood of all points on the curve except P and the origin 0. Similarly F, 1 = 0 yields the point Q with coordinates (2.y2, 2.y4). Then x is expressible as a function of yin a neighborhood of all points except Q and the origin 0. 0 PROBLEMS
In each of Problems 1 through 4 show that the relation F(x, y) = 0 yields y as a function of x in an interval I about x 0 where F(x 0 , y0 ) = 0. Denote the function by f and compute f'.
14.1. The Implicit Function Theorem for a Single Equation
347
y
Figure 14.3
2. 3. 4.
=y
+ y - x 2 = 0; (x 0 , y0 ) = (0, 0). F(x, y) = x 213 + y 213 - 4 = 0; (x 0 , Yo)= (1, 3J3). F(x, y) =xy + 2 In x + 3 In y - 1 = 0; (x 0 , y0 ) = (1, 1). F(x, y) = sin x + 2 cosy - t = 0; (x 0 , y 0 ) = (n/6, 3n/2).
1. F(x, y)
3
5. Give an example of a relation F(x, y) = 0 such that F(x 0 , y0 ) = 0 and F, 2 (x 0 , y 0 ) = 0 at a point 0 = (x 0 , y0 ), and yet y is expressible as a function of x in an interval about x 0 •
In each of Problems 6 through 9 show that the relation F(x 1 , x 2 , y) = 0 yields y as a function of (x 1 , x 2 ) in a neighborhood of the given point P(x~, xg, y 0 ). Denoting this function by f, compute f. 1 and f. 2 at P. 6. F(x 1 , x 2 , y) =xi+ x~ + y 3
-
3x 1 x 2 y- 4
= 0; P(x?, x~, y0 ) = (1, 1, 2).
= eY- y 2 - xi- x~ = 0; P(x?, x~, y 0 ) = (1, 0, 0). F(x 1 , x 2 , y) x 1 + x 2 - y- cos(x 1 x 2 y) = 0; P(x?, x~, y 0 ) = (0, 0, F(x 1 , x 2 , y) =x 1 + x 2 + y- e"'" Y = 0; P(x~, x~, y0 ) = (0, t, t).
7. F(x 1 , x 2 , y) 8.
9.
=
-1).
2
10. Prove Theorem 14.2. 11. Suppose that F is a function from IR 2 to IR 1 which we write y = F(x 1 , x 2 ). State hypotheses on F which imply that x 2 may be expressed as a function of x 1 andy (extension of the Inverse function theorem). Use Theorem 14.2.
348
14. Implicit Function Theorems and Lagrange Multipliers
12. Suppose that F(x, y, z) = 0 is such that the functions z = f(x, y), x = g(y, z), and y = h(z, x) all exist by the Implicit function theorem. Show that f.1·g,1·h,1 =
-1.
This formula is frequently written oz . ox. oy = -1. ox oy oz
13. Find an example of a relation F(x 1 , x 2 , y) = 0 and a point P(x?, xg, y 0 ) such that P satisfies the relation, and F, 1 (x?, xg, y 0 ) = F, 2 (x~, xg, y 0 ) = F, 3 (x~, xg, y 0 ) = 0, yet y is a function of(xt. x 2 ) in a neighborhood of P. 14. Suppose that the Implicit function theorem applies to F(x, y) = 0 so that y Find a formula for f" in terms ofF and its partial derivatives.
= f(x).
15. Suppose that the Implicit function theorem applies to F(x 1 , x 2 , y) = 0 so that y = f(x 1 , x 2 ). Find formulas for !, 1 , 1 ; !, 1 , 2 ; !, 2 , 2 in terms ofF and its partial derivatives.
14.2. The Implicit Function Theorem for Systems We shall establish an extension of the Implicit function theorem of Section 14.1 to systems of equations which define functions implicitly. A vector x in !Rm has components denoted (x 1 , x 2 , ••• , xm) and a vector yin !Rn will have its components denoted by (y 1 , y 2 , ... , Yn). An element in !Rm+n will be written (x, y). We consider vector functions from !Rm+n to !Rn and write F(x, y) for such a function. That is, F will have components F 2 (x, y), ... , P(x, y)
with each pi a function from !Rm+n to IR 1. In order to establish the Implicit function theorem for systems we need several facts from linear algebra and a number of useful inequalities. We suppose the reader is familiar with the elements of linear algebra and in the next three lemmas we establish the needed inequalities. Definition. Let A be an m x n matrix with elements
The norm of A, written IAI, is defined by
IAI =
L~ i~ (aj)
2
J'
2 •
Observe that for a vector, i.e., a 1 x n matrix, the norm is the Euclidean length of the vector.
349
14.2. The Implicit Function Theorem for Systems
w'
C2 , ..• ' en) Lemma 14.1. Let A be an m X n matrix, and suppose that ' = and that components n with matrix) 1 x n an is, (that vector is a column 2 that such components m with vector column a is 1'/m) 11 = (1'/1, 11 , ••• , or equivalently
(14.7)
i = 1, 2, ... , m. Then
(14.8) PROOF. For fixed i in (14.7) we square both sides and apply the Schwarz inequality (Section 6.1 ), getting
Then (14.8) follows by summing on i and taking the square root.
0
The next lemma shows that with the above norm for matrices (and vectors) we can obtain an inequality for the estimation of integrals which resembles the customary one for absolute values. Lemma 14.2. Let b: !Rm---+ !Rn be a continuous vector function on a bounded, closed figure H in !Rm. Suppose that Cis the n x 1 column vector defined by
C=
L
bdVm.
That is, i
where bi: !Rm---. IR 1, i
= 1, 2, ... , n,
= 1, 2, ... , n are the components of b.
Then
We apply Lemma 14.1 and note that since IA.I = 1 we obtain
1(1
~
L
IA.IIbl dVm =
L
lbl dVm.
0
14. Implicit Function Theorems and Lagrange Multipliers
350
Definitions. Let G be an open set in !Rm+n and suppose that F: G --+ IR" is a vector function F(x, y) with continuous first partial derivatives. We define the n x m and then x n matrices VxF and VyF by the formulas
_!__pt axl
_!__pl
_!__pl
axm
ayl VyF
VxF=
~Ft ay"
=
_!_pn
_!_pn
_!_pn
axl
axm
ayl
~F" ay"
The Fixed point theorem of Chapter 13 will be used to establish the Implicit function theorem for systems. We note that in proving this theorem for a single equation we made essential use of the Mean-value theorem. The next lemma provides an appropriate generalization to systems of the Mean-value theorem. Lemma 14.3. Let G be an open set in !Rm+n and F: G --+ IR" a vector function with continuous first partial derivatives. Suppose that the straight line segment L joining (.X, y) and (x, y) isinG and that there are two positive constants M 1 , M 2 such that
IVxFI ~ M 1
and
IVyFI ~ M2
for all points (x, y) on the segment L. Then
y)- F(x, .Y)I
IF(x, PROOF.
(x
~ M1
·lx- xl + M2 ·ly- .YI-
Any point on the segment joining (x, y) to (x, y) has coordinates ~ 1. We define the vector function
+ t(x- x), y + t(y- y)) for 0 ~ t f(t)
= F(x + t(x - x), .v + t(y -
and use the simple fact that f(1)- f(O)
=
Il
.vn
f'(t) dt.
Since f(1) = F(x, y), f(O) = F(x, y), it follows that F(x,
y) -
F(x, .Y) =
f
:t F(x + t(x- x),
.v + t(y- .Y)) dt.
Carrying out the differentiation with respect to t, and using the Chain rule, we find for each component Fi,
=
I
1 {
0
} f a . _ a . _ ~ ~ ~(F')(xi- xi)+ ~ ~(F')(.Yk- .Yd dt. j=l
uxi
k=l
uyk
14.2. The Implicit Function Theorem for Systems
351
In matrix notation we write F(x,
t ~t
n - F(x, .Y) =
[VJ · (x - x) + vyF · Cv - .Y)J dt.
From Lemma 14.2, it is clear that
IF(x, y)- F(x, .Y)I
~
[IVJI·Ix- xl + IVyFI·Iy- .YIJ dt
M1 ·lx- xl + M2 ·I.Y- H
D
For later use we next prove a simple proposition on nonsingular linear transformations. Lemma 14.4. Let B be an n x n matrix and suppose that IBI < 1. Define A = I - B where I is the n x n identity matrix. Then A is nonsingular.
Consider the mapping from IR" to IR" given by y =Ax, where x E IR", y E IR". We show that the mapping is 1-1 thereby implying that A is nonsingular. Let x 1 , x 2 E IR"; we have PROOF.
and Therefore
IAxl- Ax2l;;,: lx1- x2l -IBxl- Bx2l;;,: lx1- x2l -IBI·Ixl- x2l ;;,: lx1- x21(1 -IBI). We conclude that if x 1 =F x 2 then Ax 1 =F Ax 2 and so the mapping is D one-to-one. The next lemma, a special case of the Implicit function theorem for systems, contains the principal ingredients for the proof of the main theorem. We establish the result for functions F: IRm+n -+ IR" which have the form F(x, y) = y - Cx - 1/J(x, y)
where C is a constant n x m matrix and t/1 is such that it and its first partial derivatives vanish at the origin. Note the relation of this form ofF with the second proof of the Implicit Function theorem for a single equation given in Theorem 14.1. Although the proof is lengthy, the reader will see that with the aid of the fixed point theorem of Chapter 13 and Lemma 14.3 the arguments proceed in a straightforward manner. Lemma 14.5. Let G be an open set in !Rm+n which contains the origin. Suppose that t/1: G -+ IR" is a continuous function with continuous first partial derivatives
14. Implicit Function Theorems and Lagrange Multipliers
352
in G and that
t/1(0, 0) = 0,
Vyt/1(0, 0) = 0.
Vxt/1(0, 0) = 0,
(14.9)
Suppose that C is a constant n x m matrix, and define the function F: G -+ IR" by the formula F(x, y) = y - Cx - t/J(x, y). For any positive numbers rands, denoted by Bm(O, r) and B"(O, s) the balls in IRm and IR" with center at the origin and radii r and s, respectively. Then (a) There are (sufficiently small) positive numbers h and k with Bm(O, h) x Bn(O, k) in G and such that for each x E Bm(O, h) there is a unique element
y
E
Bn(O, k) whereby F(x, y) = 0
or, equivalently,
y = Cx
+ t/J(x, y).
(b) If g denotes the function from Bm(O, h) to B"(O, k) given by these ordered pairs (x, y), then g is continuous on Bm(O, h) and all first partial derivatives of g are continuous on Bm(O, h). PROOF
(a) Since G is open and t/1 is continuous on G, there is a positive number k such that the closed set B = Bm(O, k) x Bn(O, k) is contained in G with t/1 continuous on B. Also, because of(14.9) and the fact that the partial derivatives of t/1 are continuous, k can be chosen so small that
We fix x in Bm(O, k) and define the mapping T from B"(O, k) into IR" by the formula 1 (14.10) T(y) = Cx + t/J(x, y). We apply Lemma 14.3 to t/J(x, y), getting for x
E
Bm(O, k), y
lt/J(x, y)l = lt/l(x, y)- t/1(0, 0)1 ~ maxiVxt/JI·Ix- 01
~ 11xl
E
Bn(O, k)
+ maxiVyt/ll·ly- 01
+ 11yl.
Therefore, for x E Bm(O, k), y E Bn(O, k) it follows that IT(y)l ~ ICl·lxl
+ 11xl + 11YI·
(14.11)
Since C is a constant matrix there is a positive number M such that ICl ~ M. Now choose a positive number h which satisfies the inequality h < k/(2M + 1). The mapping (14.10) will be restricted to those values of x in the ball Bm(O, h). Then, from (14.10), for each fixed x E Bm(O, h) andy E Bn(O, k) we have IT(y)l ~ (M 1
+ 1)h + 1k < 1k + 1k = k;
In the second proof of Theorem 14.1 we denoted this mapping by TxY·
353
14.2. The Implicit Function Theorem for Systems
hence T maps Bn(O, k) into itself. Furthermore, for y 1 , y 2 e Bn(O, k) we find
IT(yl)-
T(y2)l
= ltfr(x, Y1)- tfr(x, Y2)l
~ tiY1- Y2i,
where Lemma 14.3 is used for the last inequality. Thus the mapping T is a contraction and the Fixed point theorem of Chapter 13 (Theorem 13.2) can be applied. For each fixed x e Bm(O, k) there is a unique y e Bn(O, k) such that y = T(y) or y = Cx + tfr(x, y). That is, y is a function of x which we denote by g. Writing y = g(x), we observe that the equation F(x, g(x)) = 0 holds for all x e Bm(O, h). (b) We show that g is continuous. Let x 1 , x 2 e Bm(O, h) and y 1 , y 2 e Bn(O, k) be such that Y1 = g(x 1), y 2 = g(x 2) or Y1 = Cx1 + tfr(xl, Y1)
and
Y2
=
Cx2 + tfr(x2, Y2).
Then We use Lemma 14.3 for the last term on the right, getting
or Hence lg(x 2)- g(x1)l ~(2M+ 1)1x2- x 1l,
and g is continuous on Bm(O, h). We now show that the first partial derivatives of g exist and are continuous. Let the components of g be denoted by g 1, g2 , ••• , gn. We shall prove the result for a typical partial derivative (ojoxp)g; where 1 ~ p ~ m. In !Rm let eP denote the unit vector in the p-direction. That is, eP has components (ef, e~, ... , e!) where e& = 1 and ef = 0 for j -# p. Fix x in Bm(O, h) and choose a positive number t0 so small that the points x + teP lie in Bm(O, h) for all t such that It I~ t 0. Now set x = x + teP and write g(x) = Cx + tfr(x, g(x)).
The ith component of this equation reads m
g;(x) =
L cjxi +
tfr;(x, g(x))
j=l
where the
cj are the components of the matrix C. Let l:lg; be defined by l:lg; = g;(x + teP) - g;(x).
Then from the Fundamental lemma on differentiation (Theorem 7.2), it follows
354
14. Implicit Function Theorems and Lagrange Multipliers
that
ag' =
-
t
c~
+
a .
+ !3 t/J'(x, g(x)) + e;(teP, llg) ux,
t
k=l
[:.a t/J 1(X, g(x)) + e:(teP, flg)J flg", uyk t
where e;(p, u) and e:(p, u) are continuous at (0, 0) and vanish there. Taking the definition of the vector eP into account, we can write the above expression in the form llg'
m
m
a .
m
.
-t- = j~ cJef + j~ axj 1/J'(x, g(x))ef + j~ ej(teP, llg)ef
a . g(x)) + -·e;(teP, flg)Jag" + fL.. [ !31/J'(X, k=l
(14.12)
t
uyk
where ej(p, u), j = 1, 2, ... , m, are continuous at (0, 0) and vanish there. We define the matrices Al(t) = c + V,.t/J(x, g(x)) + e(teP, llg),
A2(t) = V,t/J(x, g(x)) + e(teP, llg)
e
where the components of e are ef and the components of are =e;, i, k = 1, 2, ... , n, j = 1, 2, ... , m. Then (14.12) can be written as the single vector equation (14.13) Define B
= I - A 2 where I is then x n unit matrix. Then (14.13) becomes Bllg = AleP. t
(14.14)
According to (14.9) we have IA 2 (0)1 ::s::; !. Therefore, by Lemma 14.4 the matrix B(O) is nonsingular. Since g is continuous on .8,.(0, h), we know that llg -+ 0 as t-+ 0. Therefore the matrices A 1 (t), A 2(t), and B(t) are continuous at t = 0. Consequently B(t) is nonsingular fort sufficiently close to zero. We allow t to tend to zero in (14.14) and conclude that the limit of llgft exists; that is, (a;ax,)g 1 exists for every i and every p. The formula lim !lg t-+0 t
= B-1 (0)A 1 (0)eP
shows that the partial derivatives are continuous functions of x.
D
Theorem 14.3 (Implicit function theorem for systems). Let G be an open set in !Rm+n containing the point (x, y). Suppose that F: G-+ !Rn is continuous and has continuous first partial derivatives in G. Suppose that F(x, y) = 0
and det V,F(x, y)
=1=
0.
355
14.2. The Implicit Function Theorem for Systems
Then positive numbers h and k can be chosen so that: (a) the direct product of the closed balls Bm(x, h) and Bn(.y, k) with centers at x, y and radii h and k, respectively, isinG; and (b) hand k are such that for each x e Bm(x, h) there is a unique y e Bn(Y, k) satisfying F(x, y) = 0. Iff is the function from Bm(x, h) to Bn(Y, k) defined by these ordered pairs (x, y), then F(x, f(x)) = 0; furthermore, f and all its first partial derivatives are continuous on Bm(x, h). PROOF.
We define the matrices B
= VyF(x, y),
and write F in the form
+ B·(y- y) + f/J(x, y),
F(x, y) =A ·(x- x)
(14.15)
where ¢J is defined 2 by Equation (14.15). It is clear that ¢J has the properties f/J(x, .Y)
= o,
Vyf/J(x, .Y) =
o.
By hypothesis, Bis anonsingularmatrix. Wemultiply(14.15) by B- 1, getting B- 1 F
= B- 1 A· (x -
x)
+ (y -
y)
+ B- 1 ¢J(x, y).
Now we may apply Lemma 14.5 with B- 1 Fin place ofF in that lemma, x - x in place of x; also, y- yin place of y, - B- 1 A in place of C, and B- 1¢J in place of 1/J. It is simple to verify that all the hypotheses of the lemma are fulfilled. The theorem follows for B- 1 F. Since B- 1 is a constant nonsingular matrix, D the result holds for F. Remarks. The first partial derivatives of the implicitly defined function f may be found by a direct computation in terms of partial derivatives of F. To see this suppose that F has components F 1, F 2 , ... , pn and that f has components f 1, f 2 , ••• , fn. We write
(14.16) where Y; = Ji(x 1 , x 2 , ••• , xm). To find the partial derivatives of p, we take the derivative of pi with respect to xP in (14.16), getting (by the Chain rule) oF;
n
oF;
ap
I--=o, -+ oxp k=1 oyk oxp
i = 1, 2, ... ,
n, p = 1, 2, ... , m.
(14.17)
Treating ofkjoxP (for fixed p) as a set of n unknowns, we see that the above equations form an algebraic system of n equations in n unknowns in which, by hypothesis, the determinant of the coefficients does not vanish at (x, y). Therefore by Cramer's rule the system can always be solved uniquely. 2
In the second proof of Theorem 14.1, the function
qJ
is defined by:
F: 2 {x0 , y 0 )1/J(x, y).
14. Implicit Function Theorems and Lagrange Multipliers
356
1. Let F(x, y) be a function from IR 4 to IR 2 given by
EXAMPLE
F 1 (x1> x 2 , Yt> Y2) =X~ -X~ F 2 (xl> x 2 , y 1 , y2 ) = 2x 1 x 2
Y~
-
+ Y~ + 4,
+ x~- 2y~ + 3y~ + 8.
Let P(x, y) = (2, -1, 2, 1). It is clear that F(x, y) = 0. Verify that det VyF(x, y) ::#; 0 and find the first partial derivatives of the function y = f(x) defined implicitly by F at the point P. Solution. We have
oF
1
~= -3y~. uyl
At P, we find
Also,
oF
1
~=2x 1 ,
ux 1
Substituting the partial derivatives evaluated at Pin (14.17) and solving the resulting systems of two equations in two unknowns first with p = 1 and then with p = 2, we get
ap EXAMPLE
13
7
oxl - 32'
oxl = 16'
2. Given F: IR 5
-+
D
IR 3 defined according to the formulas
+ 2x~ x 1 + 3x 2 -
+ 4YtY2- y~ + y~, 4x 1 x 2 + 4y~- 2y~ + y~,
F 1 (x 1 , x 2 , y 1 , y 2 , y 3 ) = x~ F 2 (xl> x 2 , Yt> Jl, y3 ) =
3y~
F 3 (x 1 , x 2 , y 1 , y 2 , y 3 ) =X~ -X~ + 4y~ + 2y 2
-
3y~.
Assume that P(x, y) is a point where F(x, y) = 0 and VyF is nonsingular. Denoting the implicit function by J, determine oJifoxi at P. Solution. According to (14.17) a straightforward computation yields
( -6y 1
of 1
+ 4Y2h+ (4y 1 ux 1
of 1 8y 1 - - 4y 2 oxl
of 2 ux 1
of 3 ux 1
2Y2h- + 3y~~ = -2xl>
ap + 2y
-
oxl
of 3 3-
oxl
= 4x 2
-
1,
We solve this linear system of three equations in three unknowns by Cramer's
14.2. The Implicit Function Theorem for Systems
357
rule and obtain expressions for of 1/ox 1 , oj2jox 1 , ojljox 1 • To find the partial derivatives off with respect to x 2 we repeat the entire procedure, obtaining a similar linear system which can be solved by Cramer's rule. We leave the details to the reader. D
Definition. Let G be an open set in ~m and suppose that F: G--.. ~n is a given vector function. The function F is of class Ck on G, where k is a nonnegative integer if and only ifF and all its partial derivatives up to and including those of order k are continuous on G. The Inverse function theorem which is a Corollary to Theorem 14.1 has a natural generalization for vector functions.
Theorem 14.4 (Inverse function theorem). Let G be an open set in ~m containing the point x. Suppose that f: G--.. ~m is a function of class C 1 and that
:v = f(x),
det VJJ(x) =F 0.
Then there are positive numbers hand k such that the ball Bm(x, k) isinG and for each y e Bm(Y, h) there is a unique point x e Bm(x, k) with f(x) = y. If g is defined to be the inverse function determined by the ordered pairs (y, x) with the domain of g consisting of Bm(Y, h) and range of gin Bm(x, k), then g is a function of class C 1• Furthermore, f[g(y)] = y for y E Bm(Y, h). PRooF. This theorem is a corollary of Theorem 14.3 in which F(y, x) = y - f(x).
D
Remarks. The Inverse function theorem for functions of one variable (Corollary to Theorem 14.1) has the property that the function is one-to-one over the entire domain in which the derivative does not vanish. In Theorem 14.4, the condition det V,J =F 0 does not guarantee that the inverse (vector) function will be one-to-one over its domain. To see this consider the function f: ~ 2 --.. ~ 2 given by
f
l _
-
x2
1-
x2
(14.18)
2•
=
with domain the annular ring G {(x 1 , x 2 ): r 1 -1. 12. y 1 = x 1 cos(nx 2 /2), y 2 = x 1 sin(nx 2 /2), x 1 > 0, -1 < x 2 < 1. 13. Given the function f: IR 3 -> IR 3 where f 1 = ex 2 cos x 1 , f 2 = ex 2 sin x 1 , and f 3 =
11. y 1
2- cos x 3 . Find the points P(x 1 , x 2 , x 3 ) where the Inverse function theorem holds. 14. Given the function f: IR 2 -> IR 2 and suppose that the Inverse function theorem applies. We write x = g(y) for the inverse function. Find formulas for iJgijiJyi, i, j = 1, 2 in terms of partial derivatives of f 1 and p. Also find a formula for 8 2 9 1 ;ay~. 15. Given F: IR 4
->
IR 2 and suppose that F(x, y)
= 0 for all x = (x 1 , x 2 ) and y =
14.3. Change of Variables in a Multiple Integral
359
(y 1 , y 2 ). State conditions which guarantee that the equation oyl ox2
oy2 ox2
oxl oyl
oxl oy2
--+--=0 holds.
14.3. Change of Variables in a Multiple Integral For functions of one variable, an integral ofthe form
f f(x) dx can be transformed into
f f[g(u)]g'(u) du by the "change of variable" x = g(u), dx = g'(u) du. Such transformations are useful in the actual evaluation of many integrals. The corresponding result for multiple integrals is more complicated and, in order to establish the appropriate formula for such a change of variables, we employ several results in linear algebra. In this section we assume the reader is familiar with the basic facts concerning matrices and linear transformations. Definition. Let G be an open set in IRm and let f: G- IRm be a C 1 function. That is,fhas componentsf\f 2 , ... ,fm andfi: G -IR 1 are C 1 functions for i = 1, 2, ... , m. The Jacobian off is them x m matrix having the first partial derivative f.~ as the entry in the ith row and jth column, i, j = 1, 2, ... , m. We also use the terms Jacobian matrix and gradient, and we denote this matrix byVf
In the next theorem we restate for vector functions the Fundamental lemma of differentiation (Part (a)) and the Chain rule (Part (b)). In Part (c) we give an extension to vector functions of Equation (14.4), the formula for the derivative of the inverse of a function. Theorem 14.5. Let G and G1 be open sets in IRm with x a point in G. Let f: G- G1 be a C 1 function and denote f = (f 1 ' j2' ... 'fm). (a) We have the formula (Fundamental Lemma of Differentiation) fi(x
+ h) -
P(x) = VPCx)h m
=
I
j=l
f.~hi
+ ei(h)
+ ei(h),
i = 1, 2, ... , m, (14.19)
14. Implicit Function Theorems and Lagrange Multipliers
360
where h = (h 1 , h 2 , ••• , hm) and e(h) = (e 1 (h), ... , em( h)) are vectors, and lim 1h 1-o e;(h)/lhl = 0. (b) Let g: G1 -+ !Rm be of class C 1 and define F(x) = g[f(x)] for x E G. Then we have the Chain Rule:
(14.20)
VF(x) = Vg[f(x)] · Vf(x).
(c) Suppose that f is one-to-one with det Vf(x) # 0 on G. Then the image f( G) = G0 is open and the inverse function g 1 = f- 1 is one-to-one on G0 and of class C 1 . Furthermore, Vg 1 [f(x)] = [Vf(x)]- 1
det([Vf(x)]- 1 ) # 0 for
with
x
E
G (14.21)
or
PROOF
(a) Formula (14.19) follows directly from the Fundamental lemma of differentiation for functions in !Rm as given in Theorem 7.2. (b) Formula (14.20) is a consequence ofthe Chain rule for partial derivatives as stated in Theorem 7.3. Each component ofVF may be written (according to Theorem 7.3) m
F,;j(x) =
L g:k[f(x)] · !,~(x), j=1
which is (14.20) precisely. (c) Since f is one-to-one, it is clear that f- 1 is a function. Let y E G0 where G0 is the image of G and suppose f(x) = y. From the Inverse function theorem, which is applicable since Vf(x) # 0, there are positive numbers h and k such that the ball B(x, k) is in G and also such that for each y E B(y, h) there is a unique x E B(x, k) with the property that f(x) = y. We define g 1(y) to be the function given by the pairs (y, x). Then g 1 is of class C 1 on B(y, h) and the domain of g 1 contains B(y, h). Hence for each yin G0 , there is a ball with y as center which is also in G0 . We conclude that G0 is open. Formula (14.21) D follows from (14.20) and the Inverse function theorem. In establishing the change of variables formula we shall see that an essential step in the proof is the reduction of any C 1 function f into the composition of a sequence of functions which have a somewhat simpler character. This process can be carried out whenever the Jacobian off does not vanish. ••• , im) be a permutation of the numbers (1, 2, ... , m). A linear transformation r from !Rm into !Rm is simple if r has the form
Definition. Let (i 1 , i2 ,
The next lemma is an immediate consequence of the above definition.
361
14.3. Change of Variables in a Multiple Integral
Lemma 14.6. The product of simple transformations is simple and the inverse of a simple transformation is simple.
If / 1 and / 2 are functions on !Rm to !Rm such that the range of / 2 is in the domain of / 1, we use the notation / 1 o / 2 for the composition / 1[f2 (x)] of the two functions. The next lemma gives the precise reduction of a function on !Rm as the composition of functions each of which has an essentially simpler character. Lemma 14.7. Let G be an open set in !Rm, x E G, and let f: G--+ !Rm be a C 1 function with det Vf(x) =F 0. Then there is an open subset G1 of G containing x such that f can be written on G1 as the composition of m + 1 functions f =gm+1 og mo· .. og 1·
(14.22)
The first m functions g 1, g 2 , .•• , gm are each defined on an open set G; in !Rm with range on an open set in !Rm such that g;: G; --+ Gi+ 1, i = 1, 2, ... , m. Moreover, the components (gf, gf, ... , g;") of g; have the form g{(x 1 ,
X 2 , ••• ,
Xm) =xi
for
j =F i
gf = 0 be given, and let A: {F1 , F2 , ••• , Fn} be a subdivision of F. Choose e1 e F1, i = 1, 2, ... , n. Then
=
It 1
K[f(ei)] IJ(ei)l Vm(J'i)-
L
K[f(x)] IJ(x)l dVml
0. 6. K (x 1 , x 2) = xi + x~. F is the region in the first quadrant bounded by xi - x~ = 1, xi- x~ = 2, x 1 x 2 = 1, x 1 x 2 = 2. The inverse mapping is: u1 =xi- x~, u2 =
2x 1 x 2. 7. Prove Lemma 14.8. 8. Complete the proof of Lemma 14.10. 9. Evaluate the integral
L
x 3 dV3 (x)
by changing to spherical coordinates: xl = p cos
m. We conclude that q e lim for every m and so q ¢ Tm for every m. Thus q
U:'=
15.1. Complete Metric Spaces
377
belongs to C and because q e B0 (p 0 , r0 ), which is an arbitrary ball, the result is established. D
Corollary. A complete metric space S is of the second category. PRooF. If S were of the first category its complement would be empty, contradictng Theorem 15.2. D
By an argument similar to that used in Theorem 15.2, the next result known as the Baire Category Theorem is easily established. We omit the details.
Theorem 15.3 (Baire category theorem). A nonempty open set in a complete metric space is of the second category. We recall that a mapping f from a metric space S1 with metric d 1 to a metric space S2 with metric d2 is uniformly continuous on S1 if and only if for every e > 0, there is a b > 0 such that d2 (f(p), f(q)) < e whenever d 1 (p, q) < b, and the number b depends only one and not on the particular points p and q. With the aid of Lemma 15.1 which follows it is not difficult to show that a function which is uniformly continuous on a set A in a metric space can be extended to A, the closure of A, in such a way that f remains uniformly continuous. Moreover, the extension is unique.
Lemma 15.1. Suppose that A is a subset of a metric space S. (a) Then any point p e A- A is a limit point of A. A, then there exists a sequence {p,} c A such that p, --+ p as n --+ oo. (c) Suppose that f: A--+ S2 is uniformly continuous with S2 a metric space. If {p,} is a Cauchy sequence in A, then {f(p,)} is a Cauchy sequence in S2 • (b) If p e
PROOF
(a) Let A' be the set of limit points of A. Then by definition, A- A = {p: p e (Au A')- A}, and (a) is proved. (b) If peA- A, then p is a limit point of A and the result follows from Theorem 6.4. If p e A, we choose p, = p for all n. (c) Let e > 0 be given. From the uniform continuity, there is a b > 0 such that d 2 (f(p'), f(p")) < e whenever d 1 (p', p") < b. Since {p,} is a Cauchy sequence, there is an integer N such that d 1 (p,, Pm) < b for m, n > N. Hence d2 (f(p,), f(pm)) < e for n, m > N. Thus {f(p,)} is a Cauchy sequence. D
Theorem 15.4. Let S1 , S2 be metric spaces and suppose that S2 is complete. Let A be a subset of S1 and f: A--+ S2 be a mapping which is uniformly continuous on A. Then there is a unique mapping f*: A--+ S2 which is uniformly continuous on A and such that f*(p) = f(p) for all peA.
378
15. Functions on Metric Spaces; Approximation
PRooF. For p e A, we define f*(p) = f(p). If p e A- A, let {Pn} be a particular sequence of elements of A such that Pn -+ p as n -+ oo. Such a sequence exists according to Lemma 15.1. Since {Pn} is a Cauchy sequence, it follows from Lemma 15.1, Part (c), that {f(pn)} is a Cauchy sequence. Since S2 is complete, we define f*(p) to be the limit approached by the Cauchy sequence {f(pn) }. We show f* is uniformly continuous. Let e > 0 be given. Since f is uniformly continuous on A, there is a~> 0 such that if p', p" e A, then
d 2(f*(p'), f*(p")) < e/3
whenever d 1 (p', p")
0 be given such that x 1 x 3 ~ x 4 - ~- Then
+~
~ x2
0, fJ > 0 defined in B is a convex function. 22. In ~ 2 consider the squareS= {(x 1, x 2): 0 ~ x 1 ~ 1, 0 ~ x 2 ~ 1}. Show that the function y = g(x) = x 1 x 2 is not convex in S.
15.3. Arzela's Theorem; the Tietze Extension Theorem LetS be a metric space and !Fa family of functions from S into IR 1 . In many applications we are given such a family and wish to extract from it a sequence offunctions fn: S-+ IR 1 which converges at every point pinS to some function f If Sis an arbitrary metric space and !F is, say, any collection of continuous functions, then not much can be said about convergent sequences in /F. However, if the members of !F have certain uniformity properties which we describe below and if S is restricted properly, then it is always possible to extract such convergent subsequences (Theorem 15.20 below).
Definitions. A metric space S is separable if and only if S contains a countable dense set. It is easy to see that IRN is separable for every N. In fact, the rational points, that is, the points x:(x 1,x2, ... ,xN) such that all xi are rational numbers form a countable dense set in IRN. We now show that large classes of metric spaces are separable.
Theorem 15.19. A compact metric space S is separable.
15. Functions on Metric Spaces; Approximation
394
PROOF. We recall Theorem 6.25 which states that for every~> 0, there is a finite number of points p 1 , p2 , ••• , Pt (the number k depending on ~)such that S is contained in U~=l B(p;, ~). Consider the sequence ~ = 1, 1/2, 1/3, ... , 1/n, ... , and the totality of points which are centers of the finite number of balls of radii 1/n which cover S. This set is countable and dense, as is readily verified. D
Remark. We give an example of a space which is not separable. A sequence of real numbers x 1 , x 2 , ••• , xn, ... is bounded if there is a number M such that lx;l ~ M for all i. Denote such a sequence by x and consider the spaceS of all such elements x. It is not difficult to verify that the function (15.14)
d(x, y) =sup lx;- Y;l i
where x=(x 1 , ... ,xn•···> and y=(y 1 , ••• ,yn•···> is a metric on S. Now supposeS were separable. Then there would be a countable set {xm}, m = 1, 2, ... ,which is dense inS. We write xm = (xf, xT, ... , x::', ... ),and we construct the element y = (y 1 , y 2 , ••• , Yn• ... ) as follows:
1
Y;
= { -1
ifxf ~ 0, if xf > 0,
i = 1, 2, ....
According to (15.14) it is clear that d(xm, y) dense and S cannot be separable.
~
1 for every m. Thus {xm} is not
Definitions. LetS be a metric space and A a set inS. We denote by !Fa family of functions f: A-+ IR 1 • For p0 e A, we say that !F is equicontinuous at p0 if and only if for every 8 > 0 there is a ~ > 0 such that lf(p)- f(Po)l
0 we may choose~ =
8
1 -
x2 l
~ Jx
to obtain (15.15).
1 -
x 2 J.
395
15.3. Arzela's Theorem; the Tietze Extension Theorem y
Figure 15.7. An equicontinuous family except at x = 1.
It is important to have simple criteria for determining when a family of functionsisequicontinuous. Let/= {x: a:::;; x:::;; b} andsuppose§"isafamily of functions f: I~ IR 1. If all the functions are differentiable and the first derivative is bounded for all x E I and f E !IF, then the family is equicontinuous. To see this let lf'(x)l :::;; M for all x E I and all f E §". From the Mean-value theorem it follows at once that lf(xd- f(x2)1 :::;; lf'(OIIxl - x2l :::;; Mlx1 - Xzl·
Choosing b = 6/M, we see that the definition of equicontinuity is satisfied. For a more general criterion see Problem 3 at the end of this section.
Definition. Let§" be a family offunctions from a metric spaceS into IR 1. The family§" is uniformly bounded if there is a number M > 0 such that lf(p)l :::;; M for all p E S and all f E §". We observe that it is possible for a family to be equicontinuous and not uniformly bounded or even bounded. Perhaps the simplest example is the sequence of functions fn(P) = n, for n = 1, 2, ... , and for all pin a metric space S. The functions fn(x) = (1/n)x for x E IR 1 provide another example of an unbounded equicontinuous family as does the sequence f..(x) = n sin(x/n) given above. The next technical lemma is needed for the proof of the existence of convergent subsequences given in Theorem 15.20.
Lemma 15.5. Let A be a compact set in a metric space S. Let §" be a family of functions from A into IR 1 which is equicontinuous and uniformly bounded on A. We set D = supp,qeA d(p, q). (Dis called the diameter of A.) We define K = sup lf(p) - f(q)i, p,qeA
fe§.
15. Functions on Metric Spaces; Approximation
396 y
l/l(x)
K
l/l(x) = K for x ;;?> D.
D
Figure 15.8
Then there is a function tjl: I-+ R 1 where I= {x: 0 < x < oo} such that: (i) t/1 is nondecreasing; (ii) t/J(x) has the constant value K for all x ~ D; (iii) lf(p)- f(q)l ~ t/J(x) whenever d(p, q) ~ x,for allf E ~; (iv) limx...o+ t/J(x) = 0. (See Figure 15.8.) PRooF. We use (iii) to define t/J(x) = sup lf(p) - f(q)l where the supremum is taken for all p, q E A with d(p, q) ~ x, and for all f E ~- Since ~ is uniformly bounded, t/1 is well defined. Clearly t/1 is a nondecreasing function of x and so (i) holds. Since d(p, q) ~ D for all p, q E A, the function t/J(x) has exactly the value K for x ~ D. Thus (ii) is established. Property (iii) follows from the definition of t/J. To show (iv) holds, let 8 > 0 be given. Since ~ is equicontinuous, each point p E A is the center of a ball B(p, r) of radius r such that for any q E A is this ball, we have
lf(q) - f(p)l
~
8
2
for all
fe
~-
That fact that A is compact now allows us to use the Lebesgue lemma (Theorem 6.27). We conclude that there is a positive number() such that every ball B(q, fJ) with q E A must lie in a single one of the balls B(p, r) defined above. Suppose p0 , q0 are any points of A with d(p 0 , q 0 ) < fJ. Then p0 , q0 E B(p0 , fJ), and so p0 , q0 E B(ji, f) for some ji and f. Therefore lf(Po) - f(qo)l ~ lf(Po) - f(ji)l
+ lf(ji) -
8
8
f(qo)l ~ 2 + 2 =
8
for allfe ~Hence tjJ(fJ)
0 such that lgt(q 0 ) - g1(q 0 )1 < e/3 if k, l > N (because gk converges at all points of A). Therefore {gk(p)} is a Cauchy sequence for all p e S. The limit function f is now defined on S and we show that it is continuous. Let p0 e Sande > 0 be given. From the equicontinuity, it follows that igk(P)- gt(Po)l ~ e whenever d(p, p0 ) < ll, and this holds for all k. However, gt(P)--. f(p) and gt(P0 ) --. f(p 0 ) as k --. oo. The inequality holds in the limit and f is continuous at p 0 • (ii) We establish the uniform convergence when Sis compact. Let e > 0 be given. From Lemma 15.5 there is an x > 0 so small that lgn(p) - gn(q)l ~ 1/J(x) =
f.
3
whenever
d(p, q)
~
x,
and this inequality holds for all n. Consider all balls of radius x in S. Since S
15. Functions on Metric Spaces; Approximation
398
is compact, there is a finite number of such balls with centers say at p 1 , p2 , ... , Pm such that U:"=t B(p;, x) ::> S. We know that limn....oo gn(P;) = f(p;), i = 1, 2, ... , m. Choose an N so large that for all
n ~ N, i = 1, 2, ... , m.
Let p be any point of S. Then pis in one of the covering balls, say B(p;, x). Consequently,
provided that n > N. Since the choice of N does not depend on the point p chosen, the convergence is uniform. Since the space S is compact, the limit function f is uniformly continuous.
D Remarks (i) If fF is a family of functions defined on a metric space S with range in JRN, equicontinuity may be defined in a way completely analogous to that given when N = 1. We simply interpret the quantity lf(p) - f(q)l to be the distance in JRN. The proofs of Lemma 15.5 and Arzela's theorem hold with the modifications required when replacing distance in IR 1 by distance in IRN. (ii) Let S be any compact, separable metric space and let C(S) be the space of continuous real-valued bounded functions on S. We define
d(f, g)
= sup lf(p) -
g(p)l
peS
and it is easily verified that C(S) is a metric space. Then Arzela's theorem states that an equicontinuous, closed bounded set in the metric space C(S) is compact. If C(S, JRN) denotes the metric space of bounded continuous functions f: S ~ JRN, a similar statement holds on the compactness of bounded, equicontinuous sets. Let f: I~ IR 1 be a continuous function with I= {x: a~ x ~ b}. We may define g(x)
={
f(a)
for x b.
Clearly, g is continuous on all of IR\ coincides with f for x e I, and maxxe R'lg(x)l = maxxei lf(x)l. This example is typical of the fact that continuous functions defined on a closed subset of a metric space may be extended as a continuous function on the entire space in such a way that the supremum of the extended function does not exceed that of the original function.
15.3. Arzela's Theorem; the Tietze Extension Theorem
399
p
Figure 15.9
In the above example it is essential that the interval I is closed. The function f(x) = 1/(x- a)(x- b) is continuous on J = {x: a < x < b}, and there is no way of extending fin a continuous manner beyond this open interval.
Definition. Let A be a set in a metric space S. The distance of a point p from A, denoted d(p, A), is defined by the formula d(p, A)
= inf d(p, q). qeA.
The following useful lemma shows that the distance function we just defined is continuous on the entire metric space.
Lemma 15.6. Let A be a set in a metric space S. Let qJ(p) denote the distance from p to A. Then qJ is continuous on S. In fact, i((J(p)- qJ(q)l ~ d(p, q) for p, q E S. (See Figure 15.9.) For any e > 0 there is a point p0 From the triangle inequality we find
PROOF.
d(q, Po)
E
A such that qJ(p) > d(p, p0 ) -e.
+ d(p, Po) ~ d(p, q) + qJ(p) + e. ~
d(q, p)
From the definition of qJ(q), we get
qJ(q)
~
d(q, p0 )
~
d(p, q)
+ qJ(p) + e,
and therefore
qJ(q) - qJ(p)
~
d(p, q)
Since e is arbitrary, it follows that qJ(q) - qJ(p) q, we get the result.
+ e. ~
d(p, q). Interchanging p and D
Lemma 15.7. Let A and B be disjoint closed sets in a metric space S. Then there is a continuous function 1/J: S-+ IR 1 such that 1/J(p) = 0 for pEA, 1/J(p) = 1 for p E B, and 1/1 is between 0 and 1 for all p E S.
15. Functions on Metric Spaces; Approximation
400
PROOF.
Define
d(p, A) 1/J(p) = d(p, A)+ d(p, B)' and it is immediate that 1/J has the desired properties.
0
Remark. Let a > 0 be any number. Then the function 1/1 1(p) = -a
+ 2ai/J(p)
has the property that 1/11(p) = -a for pEA, 1/1 1(p) =a for p E Band 1/1 1 is between -a and a for all pES. Also, 1/1 1 is continuous on S. In Section 15.4 we establish several theorems which show that continuous functions may be approximated uniformly by sequences of smooth functions. In order to make such approximations it is important to be able to enlarge the domain of the continuous function. The next result gives the basic extension theorem for continuous functions. Theorem 15.21 (Tietze extension theorem). Let A be a closed set in a metric space S and f: A--+ ~ 1 a continuous, bounded function. Define M = suppeAif(p)l. Then there is a continuous function g: S--+ ~ 1 such that g(p) = f(p) for pEA and lg(p)l :::;; M for all pES.
We shall obtain gas the limit of a sequence g 1, g2, ... , gn, ... which will be determined by sequences f 1, f 2, ... , fn, ... and 1/J 1, 1/12, ... , 1/Jn, ... which we now define. We begin by setting f 1(p) = f(p) for pEA, and we define the sets A 1 = {p: pEA and f 1(p):::;; -1M}, PROOF.
B1 = {p: pEA
and f1(p);;.: 1M}.
The sets A 1, B1 are disjoint and closed. Now according to the Remark following Lemma 15.7, there is a function 1/11(p), continuous on Sand such that 1/11(p) = -1M 1/11(p) = 11/11(P)I:::;;
on A 1,
1M on B1, 1M on S.
We define
f2(p) = f1(p)- 1/11(p) for pEA. We show that lf2 (p)l :::;; (2/3)M for pEA. To see this, let p E A 1. Then f 2(p) = (1/3)M + f 1 (p). But -M:::;; f 1(p):::;; -(1/3)M on this set. Hence l(f2 (p)l:::;; (2/3)M. Now let p E B1. Then f 2(p) = f 1(p)- (1/3)M. From the definition of B1, we get lf2 (p)l:::;; M- (1/3)M = (2/3)M. Similarly, if pEA- A1 - B1,
401
15.3. Arzela's Theorem; the Tietze Extension Theorem
then both/1 and 1/11 are bounded in absolute value by (1/3)M. Now we define A2 B2
= {p: peA and / 2 (p) ~ -i·jM}, = {p: peA and / 2 (p) ~ i·jM}.
The sets A 2 , B2 are disjoint and closed. In the same way as before, there is a function 1/12(p) such that 1/12(p) = -i·jM on A 2 , 1/12(p)
= i·iM
11/12(P)I ~ i·iM
on B2 , on S.
We define and find Continuing in this manner, we define
-i"- 1M},
An= {p: peA
and f,(p) ~
Bn = {p: p e A
and J,(p) ~ i = !1- fn+1·
~
2/3)" M, we know that /,+ 1(p)-+ 0 as n-+ oo. Therefore gn -+
402 g=
15. Functions on Metric Spaces; Approximation
!1
=
f
as n--+ oo. Also, lg(p)l ~ tM[l
+~+
(~) 2
+ · · ·]
= M
for all pES.
0
PROBLEMS
0 ::::;; x ::::;; 1} and consider the totality offunctions f: I is bounded. Denote this space by fJI and for f, g E fJI, let
1. Let I
f
= { x:
->
IR 1 such that
d(f, g) = sup lf(x) - g(x)l. xel
Show that d(f,g) is a metric and that fJI is a metric space which is not separable. 2. Let x = (c 1 , c2 , ••• ,c., ... ) where the {c.} are real numbers such that c.-> 0 as n -> oo. Let '6'0 be the totality of all such sequences. Let y = (d 1 , d 2 , ••. , dm, ... ) and for x, y E '6'0 define d(x, y) = sup lc;- d;l. 1 ~i< 00
Show that '6'0 is a metric space which is separable. [Hint: Consider the elements of'6'0 of the form(r 1 , 0, 0, ... ),(r1 , r2 , 0, 0, ... ), etc., where r; are rational numbers.]
f: S-> IR 1 has the following property: there are positive numbers M, ex with 0 < ex ::::;; 1 such that
3. Let S be a metric space and suppose that
lf(p)- f(q)l ::::;; M[d(p, q)]" for any two points p, q in S. Then we say f satisfies a Holder condition on S with constant M and exponent ex. Show that a family:!' of functions satisfying a Holder condition with constant M and exponent ex is equicontinuous. 4. Let A be an open convex set in !RN and {f.} a sequence of functions f.: A -> IR 1 which are convex. Show that if {!.} converges at each point of A, then {!.} converges uniformly on each compact subset of A.
5. Let A be an open convex set in !RN and {f.} a sequence of functions f.: A-> IR 1 which are convex. Suppose that {f.} are uniformly bounded. Show that {f.} contains a convergent subsequence and that the subsequence converges uniformly on each compact subset of A.
6. Let I= {x: a::::;; x < oo} and let:!' be a family of functions f: I-> IR 1 which are differentiable on I. Let M 0 , M 1 be constants such that lf(x)l ::::;; M 0 , lf'(x)l ::::;; M 1 for all f E :!'. Is there always a subsequence of :!' which converges uniformly on I? 7. Let A be a bounded convex set in !RN and suppose A(o) is not empty. Let{!.} be a sequence of functions f.: A-> IR 1 such that f. E C 1 for each nand lf.(x)l ::::;; M 0 for all x E A and IVf.(x)l ::::;; M 1 for all x EA. Then a subsequence of {f.} converges uniformly on A. 8. Let I= {x: a::::;; x::::;; b} and S = {(x, y): a::::;; x::::;; b, a::::;; y::::;; b}, and suppose that K: S-> IR 1 is continuous on S. We set
f(x) =
r
K(x, y)g(y) dy
(15.17)
15.4. Approximations and the Stone-Weierstrass Theorem
403
where f and g are real-valued functions defined on I. For any family of functions ~. Equation (15.17) defines a family of functions !F. Show that if for all g e ~ we have jg(y)l ~ M for ally e I, then the family fF contains a uniformly convergent subsequence. 9. Prove Arzela's theorem (Theorem 15.20) for functions defined on a separable metric space with range in IRN, N > 1. 10. Give an example of a function defined on I= {x: a< x < b} with range in IR 1 which is continuous on I, bounded, and which cannot be extended as a continuous function to any larger set. i.e., show that the conclusion of Tietze's theorem does not hold. 11. In Lemma 15.7, suppose that the sets A and Bare disjoint but not closed. Show that it may not be possible to find a function 1/1 with the stated properties. 12. Let I= {x: a~ x < oo} and let f: I-+ IR 1 be continuous but not necessarily bounded. Show that f can be extended to all IR 1 as a continuous function. Prove the same result if the domain off is any closed set A c IR 1 • 13. In IR 2 let S = {(x 1 , x 2 ): 0 ~ x 1 ~ 1, 0 ~ x 2 ~ 1}, and define f(x 1 , x 2 ) = x 1 x 2 for (x 1 , x 2 ) e S. Show explicitly how f can be extended to all of IR 2 as a continuous function with no increase in the maximum of its absolute value. 14. In IR 3 let C = {(x 1 , x 2 , x 3 ): 0 ~ x 1 ~ 1, 0 ~ x 2 ~ 1, 0 ~ x 3 ~ 1}, and definef(x) = x 1 x 2 x 3 for x e C. Show explicitly how f can be extended to all of IR 3 as a continuous function with no increase in the maximum of its absolute value.
15.4. Approximations and the StoneWeierstrass Theorem Let J, defined on an interval I= {x: a< x < b} with values in IR 1 , have derivatives of all orders at a point c in/. We recall that f is analytic at c if and only iff can be expanded in a power series of the form
f(x) =
f n=O
Jl">~c) (x n.
c)",
(15.18)
and this series has a positive radius of convergence. A function is analytic on a set A if it is analytic at each point of A. As we saw in Chapter 9, a function may possess derivatives of all orders (we say it is infinitely differentiable, and write f e C 00 ) at a point and yet not be analytic at that point. To illustrate this fact consider the function
f(x) =
{e-1/x2, 0,
X X
#;0, =0.
It is not difficult to verify that JI">(O) = 0 for n = 1, 2, ... , and therefore the power series expansion (15.18) does not have a positive radius of convergence
15. Functions on Metric Spaces; Approximation
404 y
g(x) vanishes for -oo < x ,;;;; 0 and I ,;;;; x < +oo
Figure 15.10
when c = 0. However, clearly f is analytic for all x #- 0. We now define the functions g 1 and g: -1jx2 X> 0, ' { e g1(x) = O, X~ 0, and g(x) = g 1 (x)·g 1 (1- x). Then g(x) vanishes outside the unit interval I= {x: 0 < x < 1} and is positive on I (see Figure 15.10). We set M
=
L
g(x)dx
and define
1 h(x) = M
IX g(t) dt. 0
Then h E coo and h has the properties: h(x) = 0 for x ~ 0, h(x) = 1 for x ;;;;: 1 and h(x) is nondecreasing (see Figure 15.11 ). Finally, we introduce the function k: ~ 1 ~ ~ 1 by the formula k(x) = {1- h(2x- 1) k(- x)
forx;;;;: 0, for x ~ 0.
Then we verify easily that k(x) is nonnegative everywhere, and k(x) =
g
for-t~ x ~ t, for x ;;;;: 1, and for x ~ - 1.
y
h(x) = 0 for -oo < x ,;;;; 0;
h(x) = I for I ,;;;; x < +oo
Figure 15.11
15.4. Approximations and the Stone-Weierstrass Theorem
405
y
0
-I
k(x) is a
coo function1
Figure 15.12
In fact, k is nondecreasing for x ~ 0 and nonincreasing fork~ 0 (see Figure 15.12). The function k is analytic for all values of x except x = ± 1, ±t, at which points k is a coo function. Smooth functions which are equal to 1 in a neighborhood of some point and which vanish outside a larger neighborhood of the same point form a useful tool in the problem of approximation of continuous functions by classes of smooth functions. The above functions have counterparts in any number of dimensions. We define k 2 (x, y) = k(x) · k(y) and obtain a function which has the value 1 on the square So={(x,y):-t~x~t.
-t~Y~t},
which vanishes outside the square
S1 = {(x, y): -1
~
X
~
1,
-1 ~ y ~ 1},
and is such that 0 ~ k 2 (x, y) ~ 1 for all (x, y). Furthermore, k2 has partial derivatives of all orders with respect to both x and y. In IRN, we set kN(x 1 , ••• , xN) = k(x 1 )k(x 2 ) ..• k(xN) and obtain a coo function which vanishes outside the hypercube of side 2, center at 0, and which has the value 1 on the hypercube of side 1, center at 0. For many purposes it is convenient to replace kN by a function of r = (Lf= 1 x? )112 with the same essential properties. To do this we define
0. Then with x = (x 1 , x 2 , •.. , xN), the function
: vanishes outside the ball B(O, p) and has the property that
f
q>:(x) dx = 1.
B(O,p)
We also call q>: a mollifier or sometimes a modified mollifier. Suppose that f: ~N __.. ~ 1 is a continuous function. We define the mollified function, denoted f"'·P' or simply fP, by the formula
fP
=f:(x-
n;!N
~)f(~) d~.
(15.20)
The mollified functions fP are smooth and tend to f as p __.. 0, as we show in the next theorem.
Theorem 15.22. Suppose that f: ~N __.. ~ 1 is continuous and q> is a mollifier on ~N. Then for every p > 0, the function f"'.p is of class C"" on ~N and f:(x- ~)vanishes outside a ball of radius p with center at x, the integration is over a bounded region rather than over all of ~N. Therefore we may use the rule for differentiating under the integral sign given in Section 11.1, to deduce that fP has partial derivatives of all orders. Now let R > 0 be given. Since f is uniformly continuous on the closed ball B(O, R + 1), it follows that there is a nondecreasing function 1/J(p) defined on I = {p: 0 < p ~ 1} (see Lemma 15.5) such that
1/J(p) __.. 0 for all x, y e B(O, R x e B(O, R), then
as p __.. o+
and
lf(x) - f(y)l ~ 1/J(Ix - yl)
+ 1) whenever lx- yl
lfp(x) - f(x)l =
If
q>:(x -
B(x,p)
~
I
q>:(x -
B(x,p)
~ 1/J(p)
~ p ~
f
B(x,p)
1. Accordingly, if p < 1 and
~) (f(~) ~)If(~) -
q>:(x -
f(x)] f(x)l
~) d~.
d~
d~
I
15.4. Approximations and the Stone-Weierstrass Theorem
The last inequality holds because I~ - xi ~ p. From the definition of last integral on the right has the value 1, and so lfP(x) - f(x)l
~
407
. Then G'P.Pn-+ G0 as before. PROOF.
In Corollaries 1 and 2 above we obtained sequences of coo functions in IRN which can be used to approximate an arbitrary continuous function to any desired degree of accuracy. If S is any compact metric space and if C(S) is the space of continuous functions from S into IR 1 with the usual metric, then we seek those subsets of C(S) which yield approximations to all elements of C(S). The basic property required of any approximating subset is that it "separate points" in S. Definitions. Let ffl be a subset of C(S) with the property that for each pair of distinct points p, q in S there is a function fin ffl such that f(p) 1= f(q). Then we say that the set ffl separates points inS. If ffl has the further property that for every pair of distinct points p, q E S and every pair of real numbers a, b there is a function f E ffl such that f(p) = a and f(q) = b, we say that ffl separates points in S and IR 1 .
15. Functions on Metric Spaces; Approximation
408
Lemma 15.8. Let f, g be in C(S) where Sis any metric space. Define h(p) = max(f(p), g(p)) and k(p) = min(f(p), g(p)) for all peS. Then h, k are members ofC(S). The proof of Lemma 15.8 depends on the fact that for any two real numbers a, b, the relations max(a, b) = i(a
+ b + Ia -
bl),
min(a, b) = i(a
+b-
Ia - bl)
hold. We leave the details of the remainder of the proof to the reader.
Theorem 15.23 (Stone approximation theorem). Let S be a compact metric space. Let !l' be a subset of C(S), the space of continuous functions from S into ~ 1 , which separates points inS and ~ 1 . In addition, suppose that !l' has the property that max(.{, g) and min(.{, g) belong to !l' whenever f and g do. Then any Fe C(S) can be approximated uniformly on S by functions in !l'. PROOF. Let F e C(S) be given. Suppose p, q are inS and let a = F(p), b = F(q). Then, since !l' separates points inS and ~1, there is a function g = gpq in !l' such that
Since g and F are continuous at q, for any e > 0 there is a neighborhood N(q) such that g(s) > b and F(s) < b + for s e N(q). Hence
te
te
g(s) > F(s)- e for s e N(q).
(15.21)
We fix p and obtain a function gpq and a neighborhood N(q) for each q e S. Since S is compact, there is a finite subset of such neighborhoods which covers S. We denote them With these neighborhoods we associated the functions
each of which satisfies an inequality similar to (15.21). We now define
hP = max(gpq, gpq,• ... , gpqJ· Then hP is in !l' and it follows from (15.21) that
hp(s) > F(s) - e for all s e S.
(15.22)
From the way we defined hP, it follows that hp(p) = a since gpq, (p) = gpq 2 (p) = · · · = gpq)P) =a. Because hP and Fare continuous, there is a neighborhood N(p) such that
hp(s) < F(s) + e for s e N(p).
(15.23)
We can find a function hP and a neighborhood N(p) for each point pinS and, because Sis compact, there is a finite subset N(pd, N(p 2 ), ••• , N(pn) of such
15.4. Approximations and the Stone-Weierstrass Theorem
409
neighborhoods which covers S. We set h = min(hp,, hP 2 ,
••• ,
hPJ
and, by virtue of(15.22) and (15.23), it follows that h(s) > F(s) - e for all s E S, h(s) < F(s)
+ e for all s E S.
(15.24)
That is, IF(s)- h(s)l < e for all s E S. Since e is arbitrary the result is 0 established.
Definition. Let d be a subset of C(S) such that for every two functions f, g in d: (i) a.f + pg Ed for all real numbers a., p; and (ii) f · g Ed. We then say that the set d forms an algebra of functions in C(S). If S is a compact metric space and d is any algebra of functions in C(S) which separates points of S and which contains the constant functions, then we shall show that d is dense in C(S). More precisely, we shall prove that every function fin C(S) can be approximated uniformly by polynomials fn where each fn is a polynomial in functions of d. To establish this result, known as the Stone-Weierstrass theorem, we employ two elementary facts conThe most important special case of this cerning the functions IxI and approximation theorem, due to Weierstrass, states that any real-valued continuous function defined on a closed interval I= {x: a~ x ~ b} can be approximated uniformly by polynomials. Equivalently, we say that the collection of polynomials is dense in the space C(I).
Jl+X.
Lemma 15.9. Let I = {x: -1 ~ x ~ 1}, and let 0, ({)(1) < 0. Hence there is a value s such that ({)(S) = 0. That is,
+ s(t4- t2)J = S[t1 + s(t3 - td]. On the other hand, t 2 + s(t 4 - t 2) > t 1 + s(t 3 - td for all sbetween 0 and 1 S[t2
and so we contradict the fact that Sis one-to-one. We conclude that Sis strictly D monotone. a~ x ~ b} and J = {x: c ~ x ~ d} be intervals of ~1, and suppose that f: I--. VN, g: J--. VN are continuous functions which have
Theorem 16.1. Let I= {x:
the same arc C as image. Then there is a continuous monotone function S from I onto J such that f(t) = g[S(t)] fortE I.
16. Vector Field Theory; the Theorems of Green and Stokes
416
PRooF. Since g is one-to-one and continuous, we may apply Part (d) of Theorem 6.42 to conclude that g-1 is a continuous map of the arc C onto the interval J. Since the composition of continuous functions is continuous, the functionS= g- 1f is a one-to-one continuous map of I onto J. Lemma 16.1 yields the final result. D
Definitions. A path is a class of continuous functions from intervals in IR 1 into VN, any two of which are related as in Theorem 16.1. Any function in this class is called a parametric representation (or simply a representation) of the path. If every two representations in the class are related by an increasing S(t), as given in Theorem 16.1, then the path is said to be a directed path. A representation I of a path will sometimes be identified with the path itself. That is, we write "let I be a path ... " when there is no danger of confusion. Remarks. The definitions of path and arc remain unchanged for functions with domain an interval of IR 1 and range in an arbitrary metric space. Since Theorem 6.42 holds for continuous functions with range in any metric space, the proof of Theorem 16.1 is unchanged in this more general setting. Similarly, the following results on the length of paths in VN could be extended to paths in any metric space.
Definitions. Let 1: I -+ VN be a continuous function. We define the length of the path I by the formula II
l(l) = sup
L ll(ti) i=l
l(ti-dl
(16.4)
where the supremum is taken over all subdivisions ~:a=
t 0 < t 1 < ··· < t, = b of the interval I= {x: a~ x ~ b}.
If l(l) is finite, we say the path is rectifiable. Remark. The length of a path f: I-+ V1 is given by the supremum of 1 lf(ti) - f(ti-dl where f: I-+ IR 1 is a continuous function. Recalling the definition of the total variation V,.b J, of a function defined on I = {x: a ~ x ~ b}, we see at once that l(l) = V,.b fin this case.
L7=
The next result is similar to Theorems 12.1 and 12.2.
Theorem 16.2 (a) Let 1: I -+ VN be a path, and suppose that ~:
a = To < Tl < ... < T,
=b
is a subdivision of I. Define gk as the restriction of I to the interval Ik = {t: 1k- 1 ~ t ~ T,.}. Then l(l)
= l(gl) + l(gz) + ... + l(g,).
16.1. Vector Functions on IR 1
417
(b) Let IT= {t: a~ t ~ T} and define IT as the restriction of I to IT. Define s(T) = l(IT)· Then sis a continuous function for Ton I.
The proof of Part (a) for N = 2 is similar to the proof of Theorem 12.1. Then the general result can be established by induction. We leave the details to the reader. The proof of Part (b) follows the procedure given in the proof of Theorem 12.2, and we again leave the details to the reader. Part (a) of Theorem 16.2 simply states that if a path is the union of disjoint subpaths, the total length is the sum of the lengths of the subpaths. The next result establishes the important fact that the length of an arc is independent of the parameter chosen to describe it.
Theorem 16.3. Let I and g be parametric representations of the same arc C. Then l(l)
= l(g).
PROOF. LetS be the function obtained in Theorem 16.1. Set m
Lf =
m
L ll(ti) i=l
l(ti-dl.
Lg =
L lg(ti) i=l
g(ti-1)1
(16.5)
where L\: a= t0 < t 1 < · · · < tm = b is a subdivision for I and L\': c = t 0 < t 1 < · · · < tm =dis the subdivision of g which we obtain by setting t 1 = S(t1) if S is an increasing function or by setting t 1 = S(tm_ 1) if S is a decreasing function. In the first case, we see that g(t1) = l(t 1) for all i and in the second caseg(t 1) = l(tm-J In both cases the sums L1 and Lg in (16.5) are equal. Hence
D
~=~
Theorem 16.4. Let 1: I-+ VN be a continuous function. Set l(t) =
f1 (t)el + f2(t)e2 + ··· + fN(t)eN.
Then the path I is rectifiable if and only if all the };, i = 1, 2, ... , N, are of bounded variation on I.
PRooF. Let L\: a=
t0 < t 1 < · · · < tm = b be a subdivision of I. We have
1/i(tk> - t; 0 be given. From the definition of length of a path, there is a subdivision A: a = u1 < u2 < · · · < uP = b such that 1
1(1)-
48
0 such that if A1 : a = t 0 < t 1 < · · · < tm = b is any subdivision of mesh less than{) and if ei E [ti-l• t;] for each i. then
li~ 1/'(e;)l(t;- t;-d-
r
1/'(x)l
dxl < ~8.
We may assume that A1 is a refinement of A. From the Mean-value theorem applied tO the COmponentS jl> f2, ... , fN of/, there are numbers eki in the interval (t;_ 1 , t;) such that i
= 1, 2, ... , m, k = 1, 2, ... , N.
Set At; = t; - t;_ 1 and then (16.7) lffor each fixed i the numbers eki all have the same value, say
e;. then the sum
419
16.1. Vector Functions on IR 1
on the right in (16.7) would be m
L 1/'(e;)IL\t;, i=l
(16.8)
which is a Riemann sum. Then proceeding to the limit we get (16.6). Since/' is uniformly continuous on J, it is possible to show that the sum in (16.7) yields the same limit as the Riemann sum (16.8). This is known as Duhamel's principle. To establish this result, set ei:; = 1;, k = 2, 3, ... , N, i = 1, 2, ... , m, and consider the quantity
e
For each fixed k, the uniform continuity of R shows that the above sum tends to zero as the maximum of the L\t; tends to zero. We leave the remaining details to the reader. 0 Suppose that f: IR 1
-+
VN is a smooth function; we define
s(t) =
J: 1/'(r)l dt.
(16.9)
Then s represents the "directed distance" along the curve r in IRN, where r is --"-+ the graph of the point P of the radius vector 0 P from the origin to a point P --+ in IRN. That is, r is the arc /(OP) (see Figure 16.1). Distance is measured from the point corresponding to f(a). By differentiating (16.9), we have s'(t0 ) = l/'(t0 )1 and the unit vector T(t0 ) is defined by
T =/'(to)= /'(to) 1/'1 s'(to) · The vector T is the unit tangent vector to
r
at each point of the curve. Since
s is a monotone function oft we note that f and Tare functions of s. Then it
follows that df = T. ds ·
For any two vectors c, d the scalar product c ·dis defined by the formula
c· d = lclldl cos fJ where (} is the angle between two directed line segments representing the vectors c and d having the same point as base. If c and d have the representations c = c 1 e 1 + .. · + cNeN, d = d 1 e 1 + .. · + dNeN, then
c·d= c 1 d 1
+ c2d 2 + ... + cNdN.
If c • d = 0, the vectors are said to be orthogonal We now suppose that/= ft(s)el + f2(s)e2 + ...
+ fN(s)eNisa C2 function.
16. Vector Field Theory; the Theorems of Green and Stokes
420
That is, each component ,h(s) of I has a continuous second derivative. Then since T· T = 1 for all s, we may differentiate and obtain the relation T. d T ds
+ d T. T =
0
or
T. ddTs = 0.
ds
Hence T and dTjds are orthogonal vectors. The scalar curvature " of defined by K=
r
is
~~~I·
The quantity " measures the rate of change of the direction of r with respect to the arc length. We define the principal normal N to r as the unit vector in the direction of dTjds. We have dT ds = KN.
The quantity R = 1/K is called the radius of curvature of r. If I is sufficiently differentiable, we may differentiate the relation N · N and find that dNjds is orthogonal toN. We express dNjds in the form dN ds
-=
K2N1
=1
+ PT
where N 1 is a unit vector orthogonal to the subspace determined by T and N, and K 2 is called the second curvature of r. By differentiating the relation T· N = 0, we see that P= -K. We may continue this process by computing dNtfds and obtain N- 1 scalar curvatures K, K2 , ••• , "N- 1 corresponding to theN- 1 normals N1 , ... , NN_ 1 to the curve r. The above discussion is of greatest interest in ordinary Euclidean threespace. In this case the vector N 1 is called the binormal and is usually denoted by B. The vectors T, N, and B form an orthonormal set of vectors at each point of r. This set is called the moving trihedral. The quantity K 2 is usually designated by - r, and r is called the torsion of r. It is easy to verify that r = 0 for a plane curve. We choose an orientation in three-space so that B=TxN
where x denotes the usual vector product of two vectors in V3 . Differentiation of the above formula yields dB ds
=
dN dT ds x N + T x Ts = T x ( -rB- KT)
= rN.
We obtain in this way the Frenet formulas dT ds
- = KN.
'
dN ds
-=
-KT+rB '
dB ds = +rN.
(16.10)
16.1. Vector Functions on IR 1
421
Given the function f: ~ 1
EXAMPLE.
f(t) = te 1
find T, N, B,
K,
....... v3
defined by
+ t 2 e 2 + ~t 3 e 3 ,
and 1:.
Solution. We have f'(t) = e 1 + 2te 2 + 2t 2 e 3 • Therefore 1/'(t)l = (1 + 2t 2) and s(t) = J~ 1/'(t)l dt = t + ~t 3 , where distance is measured from t = 0. We
compute
Continuing the process, we obtain dT dT ds = dt(1
+ 2t 2 t
1
= (1
+ 2t 2t
+ (2- 4t 2 )e2 + 4te 3 ].
3[
-4te 1
(1
+22t2)2
Consequently, K
and N = (1 dNjds
+ 2t 2 t
1 [ - 2te 1
=
~~~~
+ (1 -
=
2t 2 )e2
+ 2te3 ]. To obtain B we compute
= (dNjdt)(1 + 2t 2 t 1 and then formula dNjds = -KT- 1:B yields the
torsion
1:.
We leave the remaining details to the reader.
D
PROBLEMS
1. Given f:IR 1 -+VN with f=(f1 , f2 , .•. ,fN) and xi=/;(t)=ti, i= 1, 2, ... , N, eliminate the parameter by expressing x 1 , x 3 , ••. , xN in terms of x 2 • In what domain of IR 1 is this elimination valid? 2. Let f(t): I-+ V2 be given by / 1 (t) = t 2 - t, / 2 (t) = t 3 - 3t, with I = {t: -3 ~ t ~ 3}. Decide whether or not the path r in IR 2 represented by f is an arc. 3. Let S be a metric space and let I, J be intervals of IR 1 . Suppose that f: I-+ S and g: J -+ S are continuous functions which have the same arc Cas image. Prove that there is a continuous monotone function h: I-+ J such thatf(t) = g[h(t)] fortE/. 4. Let S be a metric space and f: I -+ S a continuous function with I an interval of IR 1 • Define the length of the path off State and prove the analog of Theorem 16.2 for such a function f 5. State and prove the analog of Theorem 16.3 for functions from an interval I in IR 1 to a metric spaceS. 6. Given/: I-+ V3 where I= {t: 0 ~ t ~ 2rr} and/=/1 e 1 + f 2 e 2 + / 3 e 3 withft = 3 cos t, f 2 = 3 sin t, f 3 = 2t, find the length of the arc represented by f 7. Let/: I-+ V2 where/= {t:O ~ t ~ 1} and/=/1 e 1
!2=
-1 {
Jt 0,
Decide whether or not f is rectifiable.
sin~
+ f 2 e 2 with/1 =
t > 0,
t' t = 0.
t, and
16. Vector Field Theory; the Theorems of Green and Stokes
422
8. Complete the proof of Theorem 16.5.
In each of Problems 9 through 12 find T, N, B, t for the given function f: IR 1 -+ V3 • t = 0.
9. l(t) = e'[(cos t)e 1 +(sin t)e 2 + e3 ], 10. l(t) = (1/3)t 3 e 1 11. l(t) = te 1
+ 2te 2 + (2/t)e 3 ,
and t at the given value of
K,
t = 2.
+ (3/2)t 2 e2 + (3/2)t 3 e3 ,
t = 2. t = 0.
12. l(t) = (t cos t)e 1 + (t sin t)e 2 + te 3 ,
13. Let 1: I--+ VN be the representation of a curve r in IRN. Suppose that I is a eN function. Let The the tangent vector and lets denote arc length. Then dTfds =~eN where N is orthogonal to T. Differentiate N with respect to s and obtain a unit vector N1 such that
Differentiate N 1 with respect to s and obtain a unit vector N2 such that dN1 = -K 1 N ds
-
+ K2N2.
Continue this process and obtain a sequence of N mutually orthogonal unit vectors T, N, N 1 , ••• , NN_ 2 and the formulas dNk
ds =
-KkNk-1
Finally, show that dNN_ 2jds called the curvatures of r. 14. Given 1: I--+ v3 where I (a) Prove that
+ Kk+tNk+t•
"N- 2NN_ 3 • The quantities K, K1 ,
= -
= { t:
a~ 2 _
k = 2, 3, ... , N - 3.
X ~
(f'
"N- 2 are
b} and suppose that I is a C 3 function. X
f") · (f'
f")
X
11'1 2
" -
••• ,
•
(b) Prove that f'·(f"
X
f"')
r = '--:--11::-'-,-x-1-=""'12--'-. In each of Problems 15 through 18 find the length of the given arc. 15. l(t) = te 1
+ !t 2 e2 + !t3 e3 ,
16. l(t) = te 1
+ tJ2t 2 e 2 + We 3 ,
I=
17. l(t) = t cos te 1 + t sin te 2 + te 3 , 18. l(t)
=
{t: 0 ~ t
~ 2}.
I= {t: 0 ~ t ~ 2}. I= {t: 0 ~ t ~ n/2}.
te 1 + log(sec t + tan t)e 2 + log sec te 3 ,
I= {t: 0 ~ t ~ n/4}.
19. State and prove the analog of Theorem 16.1 when the range of I and g are in a metric space.
423
16.2. Vector Functions and Fields on IRN
Figure 16.3. A vector field.
16.2. Vector Functions and Fields on IRN In this section we develop the basic properties of differential and integral calculus for functions whose domain is a subset of ~M and whose range is in a vector space VN. We shall emphasize the coordinate-free character of the definitions and results, but we shall introduce coordinate systems and use them in proofs and computations whenever convenient. Definition. A vector function from a domain D in ~M into VN is a mapping
f: D -+ VN. If the image is V1 (which we identify with ~ 1 in the natural way) then f is called a scalar function. Vector or scalar functions are often called vector or scalar fields. The term field is used because we frequently visualize a vector function as superimposed on ~M. At each point of the domain D of ~M a representative directed line segment off is drawn with base at that point. We obtain a field of vectors (see Figure 16.3 where M = N = 2). Let f: D -+ VN be a vector function where D is a domain in ~M. An orthonormal basis for VN is a set of N unit vectors e 1 , e2, ... ,eN such that ei· ei = 0 for i #- j, i, j = 1, 2, ... , N. Then for each point P ED, we may represent f by the formula N
/(P) =
L /;(P)ei i=1
(16.11)
where/;: D-+ ~1, i = 1, 2, ... , N, are scalar functions from D into ~ 1 • We can consider vector fields with domain in any space such as VM or EM (Euclidean space). Equation (16.11) illustrates what is meant in that case. We shall ordinarily assume that Dis in ~M. The function f is continuous at P if and only if each /;, i = 1, 2, ... , N, is continuous at P. We recall that the basic properties of functions from ~Minto ~ 1 were developed in Chapters 6, 7, and 8. Properties developed there which involve linear processes are easily transferred to functions f from ~M into VN. For simplicity we shall usually consider functions f from ~N into VN, although many of the results developed below are valid with only modest changes if the dimension of the domain off is different from that of its range. The definite integral of a vector function, defined below, is similar to the integral of a scalar function on ~N as given in Chapter 8.
424
16. Vector Field Theory; the Theorems of Green and Stokes
Definition. Let D be a domain in ~N and suppose that u: D --+ VN is a vector function. Let F be a figure in D. The function u is integrable over F if and only if there is a vector L in VN with the following property: for every e > 0, there is a~> 0 such that if A= {F1 , F2 , ••• , Fn} is any subdivision ofF with mesh less than ~ and p 1 , p2 , ••• , Pn is any finite set of points with P; in F;, then
Iit
I
u(p;) V(F;) - L < e,
where V(F;) is the volume ofF;. We call L the integral of u over F and we write
L
L
udV.
=
If, for example, u is given in the form (16.11) so that N
u(P)
=
L u;(P)e;,
(16.12)
i=1
then it is clear that u is integrable over F if and only if each scalar function u; is integrable over F. Therefore if L exists it must be unique. In fact, if u is integrable and given by (16.12), we have
f
F
u dV =
.f (J u; dv)e;.
&=1
(16.13)
F
Each coefficient on the right in (16.13) is an integral of a function from ~N into ~ 1 as defined in Chapter 8. Let v be a vector in VN and denote by Pc;P a directed line segment in ~N which represents v. As usual, we use the notation v(p 0 p} for this vector. Definitions. Let p0 be a point in ~N and let a be a unit vector in VN. Suppose that w: D --+ VN is a vector function with domain D in ~N. We define the directional derivative of win the direction a at p0 , denoted by D.w(p 0 ), by the formula D ( ) 1. w(p)- w(p 0 ) .w Po = tm h h-+0
where the point p e D is chosen so that v(Pc;P) = ha (see Figure 16.4). The vector function w is continuously differentiable in D if and only if wand D.w are continuous on D for every unit vector a in VN. We write we C 1 (D).
Figure 16.4
16.2. Vector Functions and Fields on IRN
425
In order to derive formulas for computing directional derivatives we require a coordinate system in IRN. When we write w(p), p E IRN we indicate the coordinate-free character of the function. If a Cartesian coordinate system (x 1 , x 2 , ... , xN) is introduced, then we write w(x 1 , x 2 , ... , xN) or w(x). If e 1 , e 2 , ••• , eN is an orthonormal basis in VN, we use coordinates to write N
w(x)
=
~ w;(x)e;, i=l
where each of the scalar functions w;(x) is now expressed in terms of a coordinate system. Strictly speaking we should use different symbols for a function w when expressed in terms of a particular coordinate system as compared with w defined in a purely geometric manner. However, it will be clear from the context whenever coordinates are used. Furthermore, we omit the statement and proofs of the theorems which show, for example, that a vector function w has a directional derivative if and only if w has one is an arbitrary Cartesian coordinate system, that w is continuous if and only if w(x) is, and so forth. The following theorem establishes the formula for obtaining the directional derivative of scalar and vector fields. Theorem 16.6 (i) Let a be a unit vector in VN. Suppose that w: D-+ IR 1 is a continuously differentiable scalar field with domain D c IRN. If a has the representation a= a 1 e 1 + a 2 e 2 + .. · + aNeN, then N
OW
D,w= ~~a;.
(16.14)
i=l UX;
(ii) If w: D-+ VN is a vector field and w = ~f=t w;e; then N
N
i=l
~j=l
D,w = ~ D,w;e; = ~ w;,iaiei.
(16.15)
PROOF. (i) If a is a unit vector in one of the coordinate directions, say xi, then (16.14) holds since D,w is the partial derivative with respect to xi. In the general case, we fix a point x 0 = (x?, ... , x2) and define
cp(t) = w(x?
+ a 1 t, ... , x2 + aNt).
Then D,w(x 0 ) is obtained by computing cp'(O) according to the Chain rule, Theorem 7.3. The proof of (ii) is an immediate consequence offormula (16.14) 0 and the representation of win the form ~ w;e;. The next theorem shows that the directional derivative of a scalar function may be expressed in terms of the scalar product of the given direction a with a uniquely determined vector field. Observe that the result is independent of the coordinate system, although coordinates are used in the proof.
426
16. Vector Field Theory; the Theorems of Green and Stokes
Theorem 16.7. Let f: D-+ IR 1 be a continuously differentiable scalar field with domain D in IRN. Then there is a unique vector field w: D-+ VN such that for each unit vector a e VN and for each p e D, the directional derivative off is given by the scalar product of a and w:
D.f(p) = a· w(p).
(16.16)
If el> e2, ... , eN is an orthonormal set of unit vectors in VN, then w(x) may be computed by the formula N
w(x) = PROOF.
L f.i(x)ei j=l
for each x e D.
(16.17)
From Formula (16.14), we find N
D.f(x) =
L f.i(x)ai. j=l
(16.18)
We now define w(x) by (16.17) and therefore the scalar product of wand a yields (16.18). To show uniqueness, assume that w' is another vector field satisfying (16.16). Then (w - w') ·a = 0 for all unit vectors a. If w - w' =I= 0, choose a to be the unit vector in the direction of w - w', in which case (w - w') · a = Iw - w'l =I= 0, a contradiction. 0 Definition. The vector w defined by (16.16) in Theorem 16.7 is called the gradient off and is denoted by grad f or Vf.
Remark. Iff is a scalar field from IRN into IR 1, then in general f(p) = const. represents a hypersurface in IRN. At any point p, a vector a tangent to this hypersurface at p has the property that Lf=t f. 1(p) · a 1 = 0 where a= Lf=t a 1e1• We conclude that Vf is orthogonal to the hypersurface f(p) = constant at each point p on the hypersurface. EXAMPLE 1. Let et> e2 , e 3 be an orthonormal set of vectors in V3(1R 3). Given the scalar and vector functions
+ x~ + x~- 3xtx2x 3 u(x1, x2, x 3) = (xf- x2 + x3)e1 + (2x2- 3x3)e2 + (xl + x 3)e3 the vector a= A.e 1 + JJ.e 2 + ve 3, A. 2 + J1. 2 + v2 = 1, find Vf and D.u f(xl, x2, x 3) = x~
and terms of x 1 , x 2 , x 3 , et> e2 , e 3 •
in
Solution. According to (16.17), we find
Vf = 3(xf- x 2x 3 )e 1 + 3(x~- x 1x 3 )e2 + 3(x~- x 1x 2)e3 • Employing (16.15), we obtain
D.w = (2x 1A.- Jl. + v)e 1 + (2Jl.- 3v)e2 +(A.+ v)e 3.
0
Theorem 16.8. Suppose that f, g, and u are C 1 scalar fields with domain D c IRN.
16.2. Vector Functions and Fields on IRN
427
Let h: IR 1 --+ IR 1 be a C 1 function with the range of u in the domain of h. Then
V(fg)
V(f +g)= Vf + Vg,
v
G) = :2
= fVg + gVf,
if g 1= 0,
(gVf - fV g)
Vh(u) = h'(u)Vu.
PROOF. We prove the second formula, the remaining proofs being left to the reader. Let e 1 , e2 , ••• , eN be an orthonormal set in VN. Then from the formula for V(fg) given by (16.17) it follows that N
V(fg)
=
L (fg),iei.
j=l
Performing the differentiations, we find N
V(fg) =
L (fg,i + f.ig)ei =
fVg
+ gVf
D
j=l
The operator V carries a continuously differentiable scalar field from IRN to IR 1 into a continuous vector field from IRN to VN. In a Cartesian coordinate system, we may write V symbolically according to the formula N
0
j=l
oxj
V= L-ei. However, the operator V has a significance independent of the coordinate system. Suppose that w is a vector field from IRN into VN which in coordinates may be written w = 1 wiei. We define the operator V · w by the formula
Lf=
NOW·
V·w=I-1, j=l
oxj
and we shall show that this operator, called the divergence operator, is independent of the coordinate system (Theorem 16.9 below).
Definitions. The support of a scalar function f with domain D is the closure of the set of points p in D where f(p) 1= 0. If the support of a function is a compact subset of D, we say that f has compact support in D. Iff is a en function for some nonnegative integer n, with compact support, we use the symbol f E C~(D) to indicate this fact. Lemma 16.2. Let D be an open figure in !RN and suppose that f
L
f.i(x) dV = 0,
E
CJ(D). Then
j = 1, 2, ... , N.
PROOF. Let S be the set of compact support of f. Then f = 0 and Vf = 0 on D - S. Let R be any hypercube which contains i5 in its interior. We extend the definition off to be zero on R - D. Then integrating with respect to xi
16. Vector Field Theory; the Theorems of Green and Stokes
428
first and the remaining N - 1 variables next, we find (in obvious notation)
f
f,j(x) dVN
= =
f [fbi
f
f,j(x) dVN
f
[f(bi, x') - f(ai, x')] dVN- 1
=
R'
R
D
f,j(x) dxi] dVN-
1
ai
= 0.
D
R'
The next lemma, useful in many branches of analysis, shows that a continuous function f must vanish identically if the integral of the product off and all arbitrary smooth functions with compact support is always zero.
Lemma 16.3. Let D be an open figure in IRN and suppose that the scalar function f is continuous on D. If
L
for all g E CJ(D),
fgdVN = 0
then f
=0 on D.
(16.19)
PRooF. We prove the result by contradiction. Suppose there is an x 0 ED such
that f(x 0 ) ¥- 0. We may assume f(x 0 ) > 0, otherwise we consider - f Since f is continuous and D is open, there is a ball in D of radius 3r0 and center x 0 on which f > 0. Denoting distance from x 0 by r, we define the function
1
g(x) = for r
0 It is easily verified that g(x)
f
D
fgdVN =
E
f
~
2r0 •
CJ(D) with g(x) ~ 0. We have
fgdV
B(x 0 , 3r 0 )
which contradicts (16.19). Hence f
~
f
fgdV > 0,
B(x0 , r 0 )
= 0 on D.
D
Remark. Lemma 16.3 remains valid if the functions gin formula (16.19) are restricted to the class C0 (D). To see this merely replace g by its mollifier as defined in Chapter 15 and proceed with the same method of proof. We now show that the divergence operator is independent of the coordinate system.
Theorem 16.9. Let D be an open figure in IRN and suppose that wE C 1(D) is a vector field. Then there is a unique scalar field v, continuous on D, such that for all u E CJ(D), we have
L
(uv
+ Vu· w) dV = 0.
(16.20)
16.2. Vector Functions and Fields on IRN
429
Furthermore, if e 1 , e2 , ••• , eN is an orthonormal basis and if, for each point xeD, w is given by N
L w1(x)e1,
w(x) =
J=l
then N
v(x) =
a
L ~ w1(x). j=l uXJ
(16.21)
PROOF. Suppose that w is given and that e 1 , ••• , eN is an orthonormal basis in some coordinate system. We define v by (16.21). Let u be any function in class CJ(D). Then, according to Lemma 16.2,
f
D
(uv
+ Vu · w) dV = =
ff
(u
D }=1
~a
uXj
w1 +
f J ~a (uw
j=l
D uXj
1)
~u w1)
dV
uXj
dV = 0.
To show that vis unique, suppose that v' is another scalar field which satisfies (16.20). By subtraction we get
L
(v - v')udV = 0
Thus v'
=v according to Lemma 16.3.
for all u e CJ(D).
0
Definition. The scalar field v determined by (16.20) in Theorem 16.9 and defined in any coordinate system by (16.21) is called the divergence of w. We use the notation v = div w or v = V · w. We note that vis determined in (16.20) without reference to a coordinate system, although (16.21) is used for actual computations. Theorem 16.10. Let D be a domain in ~N and suppose that w, w1 , w2 , ••• , Wn are C 1 vector fields on D. Let f be a C 1 scalar field on D and suppose that c 1 , c2 , ••• , en are real numbers. Then (a) div(Lj= 1 c1w) = Lj= 1 c1 div w1. (b) div(fw) = f div w + Vf· w. We leave the proof to the reader. EXAMPLE 2. Suppose that the origin of a coordinate system is at the center of the earth and R is its radius. Denote by g the acceleration due to gr!!:Ytty at the surface of the earth. For points pin ~ 3 let r be the vector having Op as a representative, and set r = lrl. From classical physics it is known that the vector field of force due to gravity, denoted v(p) and called the gravitational
430
16. Vector Field Theory; the Theorems of Green and Stokes
field of the earth, is given approximately by gR2
v(p) = - 3 r r
for
r > R.
Show that div v(p) = 0 for r > R. Solution. We use Part (b) of Theorem 16.10 to obtain
Next, we introduce the unit vectors e 1 , e2 , e3 and obtain div r = 1 + 1 + 1 = 3,
Using Theorem 16.8 and the above formulas, we find div v = - 3gR2r3
(-3gR2)~-r = r4
0.
r
0
The vector or cross product of two vectors in a three-dimensional vector space allows us to introduce a new differential operator acting on smooth vector fields. We shall define the operator without reference to a coordinate system although coordinates are used in all the customary computations. If e 1 , e 2 , e 3 is any orthonormal basis for V3 we may construct the formal operator e3 el e2 a a a Vxu= oxl ox2 OX3 u3 u2 ul = (ou 3_ ou 2)e 1+ (ou 1_ ou 3)e2+ (ou2 _ ou 1)e3. oxl oxl OX3 oxl ox2 ox3
(16.22)
We note that if the symbol V is replaced by a vector v with components v1 , v2 , v3 and if the partial derivatives in (16.22) are replaced by these components, then we obtain the usual formula for the vector product of two vectors.
Theorem 16.11. Let D be any set in IR 3 and let u e C 1 (D) be a vector field into V3 • Suppose that IR 3 is given one of its two possible orientations. Then there is a unique continuous vector field w from D into V3 such that div(u x a)= w·a
(16.23)
for every constant vector a. If e 1 , e 2 , e 3 is any orthonormal basis of a coordinate system consistent with the orientation of IR 3 , then the vector w is given by (16.22).
16.2. Vector Functions and Fields on IRN
431
PROOF. With the orthonormal basis e 1, e 2, e 3 given, we write a= a 1e 1 + a2e 2 + a 3 e 3 , u = u 1e 1 + u2e 2 + u 3 e 3 • The formula for the vector product yields
u x a= (u 2a 3 - u 3a2)e 1 + (u 3a 1 - u 1a3)e2 + (u 1a2 - u2a1)e 3. Using (16.21) for the divergence formula, we find
. ou2 ou3 dtv(u x a)= a 3 - - a2 ox1 ox1
ou3 ou1 ou1 ou2 + a 1- a3 + a2- a1- . ox2
ox2
ox3
ox3
If we denote by wl> w2 , w3 the coefficients in the right side of (16.22), then
div(u x a)= w1 a 1 + w2 a2 + w3 a 3 • That is, div(u x a)= w·a where w = w1 e 1 + w2 e 2 + w3 e 3 . The vector w is unique since if w' were another such vector we would have w ·a = w' ·a for all unit vectors a. This fact implies that w = w'. 0
Definition. The vector win Theorem 16.11 is called the curl of u and is denoted by curl u and V x u. The elementary properties of the curl operator are given in the next theorem.
Theorem 16.12. Let D be a domain in IR 3 and suppose that u, v, u1 , .•. , un are C 1 vector fields from D into V3 . Let f be a C 1 scalar field on D and c 1 , ••• , en real numbers. Then (a) (b) (c) (d) (e)
curl(Lj= 1 ciui) = Lj= 1 ci(curl ui). curl(fu) = f curl u + Vf x u. div(u x v) = v·curl u- u·curl v. curl Vf = 0 provided that Vf E C 1 (D). div curl v = 0 provided that v E C 2 (D).
PRooF. We shall establish Part (b) and leave the remaining proofs to the reader. Letel>e 2 ,e 3 be an orthonormal basis and setu = u 1 e 1 + u 2 e 2 + u 3 e 3 • Then
Therefore curl(fu) = f curl u + ( u 3 : :2 - U2
+ ( U1 ::3 -
u3
::J
e2
::J
e1
+ ( U2 0~
1 - U1 a~J e3.
(16.24)
Recalling that Vf = (offoxde 1 + (offox 2)e2 + (offox 3)e 3 and using the
16. Vector Field Theory; the Theorems of Green and Stokes
432
formula for the vector product, we see that (16.24) is precisely curl(fu) = f curl u
+ Vf
x u.
0
Remark. Part (d) in Theorem 16.12 states that if a vector u is the gradient of a scalar function J, that is if u = Vf, then curl u = 0. It is natural to ask whether or not the condition curl u = 0 implies that u is the gradient of a scalar function f. Under certain conditions this statement is valid. In any Cartesian coordinate system, the condition u = Vf becomes
of u2 = ox2'
or (16.25) Then the statement curl u = 0 asserts that the right side of(16.25) is an exact differential. EXAMPLE
3. Let the vector field u be given (in coordinates) by u = 2x 1 x 2x 3e 1
+ (xix 3 + x 2)e2 + (xix 2 + 3xDe3.
Verify that curl u = 0 and find the function f such that Vf = u. Solution. Computing V x u by Formula (16.22), we get
curl u =
et
e2
e3
a
a
a
oxl
ox 2
ox3
2x 1 x 2x 3
xfx3
=(xi- xi)e 1
+ x2
+ (2x 1 x 2 -
xfx 2 + 3x~
2x 1 x 2)e2 + (2x 1 x 3 - 2x 1 x 3)e3 = 0.
We seek the function f such that
Integrating the first equation, we find f(xl,
X2,
X3) = XIX2X3
+ C(x2, X3).
Differentiating this expression with respect to x 2 and x 3, we obtain
of
2
= XtX3 ox 2
ac +ox- = 2
Thus it follows that
ac
2
XtX3
-=x2 ' ox 2
+ X2
16.2. Vector Functions and Fields on IRN
where K'(x 3 ) stant. Hence
= 3x~.
Therefore C(x 2 , x 3 )
433
= !x~ + x~ + K 1
with K 1 a con-
0 PROBLEMS
1. Let D be a domain in IRN and suppose that u: D -+ VN is a vector function. IfF is a figure in D and if JFudVexists, prove that the value is unique.
Lf=
2. Suppose that u: D -+ VN is integrable over some figure Fin D. If u(x) = 1 u1(x)ei for some orthonormal basis e 1 , .•. , eN, write a detailed proof of the formula
3. Suppose that u: IR 2 -+ V2 is given by u(x 1 , x 2 ) = (xf- xDe 1 JFudVwhere F = {(x 1 , x 2 ): xf + x~ ~ 1}.
+ 2x 1 x 2 e 2 •
Find
In each of the following Problems 4 through 7 express Vf(x) in terms of (x 1 , x 2 , x 3 ) and(el> el> e3 ) andcomputeD.,f(x) where a is the given unit vector and xis the given point.
= 2xf + x~- x 1 x 3 - x~, a= (1/3)(2e 1 - 2e2 - e 3), x = (1, -1, 2). f(x) = xf + x 1 x 2 - x~ + x~. a= (1/7)(3e 1 + 2e 2 - 6e 3), x = (2, 1, -1). f(x) = e"• cos x 2 + e"• cos x 3, a= (1/J3)(e 1 - e2 + e 3), x = (1, 1t, -1/2). f(x) = xf log(1 + xD - x~, a = (1/.Jl0)(3e 1 + e3), x = (1, 0, - 2).
4. f(x)
5. 6. 7.
In each of Problems 8 and 9 express D.,w(x) in terms of (x 1 , x 2 , x 3 ) and (e 1 , e 2 , e3 ). Find the value at xas given. ·
= x 2 x 3 e 1 + x 1 x 3 e2 + x 1 x 2 e3 , a = (1/3)(e 1 + 2e2 - 2e3), x = (1, 2, -1). w(x) = (x 1 - 2x 2 )e 1 + x 2 x 3 e2 - (x~ - x~)e 3 , a = (1/.ji4)(3e 1 + 2e 2 - e 3), x =
8. w(x) 9.
(2, -1, 3). 10. Prove the first, third, and fourth formulas in Theorem 16.8. 11. Show that Lemma 16.3 remains valid if the functions gin Formula (16.19) are restricted to the class q'(D). 12. Prove Theorem 16.10.
In each of Problems 13 through 17 find the value of div v(x) at the given point
x. 13. v(x)
16.
xfe3 ,
x = (1, 0, 1).
= (xf- X 2 X 3)e 1 + (x~- x 1 x 3)e2 + (x~- x 1 x 2 )e3, x = (2, v(x) = Vu, u(x) = 3x 1 x~ - x~ + x 3, x = (-1, 1, 2). v(x) = r-•r, r = x 1 e 1 + x 2 e2 + x 3 e3 , r = lrl, x = (2, 1, - 2).
14. v(x) 15.
= x 1 x 2 e 1 + x~e2 -
-1, 1).
434
16. Vector Field Theory; the Theorems of Green and Stokes
In each of Problems 18 through 20, find curl v in terms of (x 1 , x 2 , x 3 ) and (e 1 , e 2 , e 3 ).1f curl v = 0 find the function f such that Vf = v. 18. v(x) =(xi + x~ + x~)- 1 (x 1 e 1 + x 2 e 2 + x 3 e 3 ). 19. v(x) =(xi + x 2 x 3 )e 1 + (x~ + x 1 x 3 )e2 + (x~ + x 1 x 2 )e 3 . 20. v(x) = e"•(sin x 2 cos x 3 e1 + sin x 2 sin x 3 e 2 +cos x 2 e3 ). 21. Find curl v where vis the gravitation field given in Example 2. 22. Prove Parts (a), (c), (d), and (e) ofTheorem 16.12. 23. Find a formula for curl(curl u) in terms of(x 1 , x 2 , x 3 ) and (el> e2 , e3 ) ifu = u1 e 1 + u 2 e 2 + u 3 e 3 and each of the u; is a C 2 function on IR 3 .
16.3. Line Integrals in
~N
Let I= {t: a~ t ~ b} be an interval and fa vector function with domain I -and rangeD, a subset of VN. We consider in ~N the directed line segments having base at the origin 0 which represent the vectors fin D. The heads of these directed line segments trace out a curve in ~N which we denote by C. In Figure 16.5, OP represents a vector in D, the range off We recall that iff is continuous and the curve C has finite length, then we say that Cis a rectifiable path (see Section 16.1). Letg be a continuous vector function from C into VN.If J, as defined above, is rectifiable then we can determine the Riemann-Stieltjes integral of g with respect to f To do so we introduce a coordinate system with an orthonormal basis el> e2, ... , eN in VN, although the formula we shall obtain will be independent of the coordinate system. We write
-
g(x) = gl(x)e 1 + ··· + gN(x)eN,
+ """ + fN(t)eN, X E C, t E I. (16.26) If /is rectifiable, then the functions fi: I--..~\ i = 1, 2, ... , N, are continuous /(t) = f1(t)el
and of bounded variation. Also, if g is continuous, then the functions gi: C--.. ~N. i = 1, 2, ... , N, are continuous. Using the notation gi[/(t)] for gi[f1 (t),
0
Figure 16.5. The curve /(t).
16.3. Line Integrals in IRN
435
... , fN(t)], we observe that the following Riemann-Stieltjes integrals exist: i = 1, 2, ... , N.
(16.27)
We now establish the basic theorem for the existence of a Riemann-Stieltjes integral of a vector function g with respect to a vector function f
Theorem 16.13. Suppose that 1: I
--+ VN is a vector fu~n and that the range C in IRN as defined above by the radius vector r(t) = v(OP) is a rectifiable path. Let g be a continuous vector field from C into VN. Then there is a number L with the following property: for every e > 0, there is a > 0 such that for any subdivision.!\: a= t 0 < t 1 < · · · < t" = b with mesh less than 0 and any choices of ei with ti-l ~ ei ~ ti> it follows that
o
Iit g[l(ei)] · [r(ti)- r(ti-d] - L I< e.
(16.28)
The number L is unique.
PRooF. We introduce an orthonormal set e 1 , e2, ... , eN in VN and writeg and in the form (16.26). Then we set
I
N fba gi[l(t)] dJi(t),
L = i~
and it is clear that each of the integrals exists. Replacing each integral by its Riemann sum, we obtain Inequality (16.28). 0
Definition. We write L = f!g[l(t)] · dl(t) and we call this number the Riemann-Stieltjes integral of g with respect to f EXAMPLE
1. Given I and g defined by
l(t) = t 2 e 1
+
2te 2
-
te 3 ,
Find f~g·df Solution. We have
L
g[l(t)]. dl(t)
Let / 1
t
1
+ t 2 ( -t)e2 + t 2 (2t)e 3 ]' [d(t 2 e 1 + 2te2 -
[(t4
-
t)e 1
= tl [(t4
-
t)·2t
=
= {t: a 1 ~ t
~
+ t 2 (-t)·2 + t 2 (2t)(-1)] dt = -;.
bd
and / 2
= {t: a2 ~ t ~ b2 }
te 3 )]
0
be any intervals and
436
16. Vector Field Theory; the Theorems of Green and Stokes
suppose that 11 : I 1 -+ C and 12 : I 2 -+ Care one-to-one mappings onto a path C in !RN, as described at the beginning ofthis section. We saw in Theorem 16.1 that there is a continuous function U: I 1 -+ I 2 such that (16.29) and that U is either increasing on I 1 or decreasing on I 1 . If C is rectifiable and g: C -+ VN is continuous, then
J.~' g[l1 (t)] · dl1 (t) = ±
J.:,
g[l2(u)] · dl2(u),
the plus sign corresponding to U increasing and the minus sign to U decreasing on I 1 • Definitions. Let 1: I-+ VN define an arc C in !RN by means of the radius vector --+ --+ r(t) = 11(0P), where OP represents f We consider the collection .!11 of all functions /,.: Ia -+ VN such that the range of fa is C and fa is related to I by an equation such as (16.29) with Ua increasing. We define a directed arc, denoted C, as the ordered pair (C, d). Any function fa in .!11 is a parametric representation of C. From Theorem 16.1 it follows that there are exactly two directed arcs c1 and c2 corresponding to an undirected arc c. We write IC11 = IC21 = c. If 11 is a parametric representation of c1 and is one of c2, then 11 and 12 are related according to (16.29) with U decreasing. It is therefore appropriate to write C1 = - C2 (see Figure 16.6).
h.
The Riemann-Stieltjes integral along a directed arc r is defined by the formula fc.g(r)·dr =
r
Cwith radius vector
g[l(t)] ·dl(t)
(16.30)
where I is a parametric representation of C. It is not difficult to see that the integral along a directed arc as in (16.30) depends only on g and Cand not on the particular parametric representation of C. Also, it follows at once that
f-c
_g(r)·dr = - i_g(r)·dr.
Jc
-Figure 16.6. Directed and undirected arcs.
16.3. Line Integrals in IRN
437
A directed arc may be decomposed into the union of directed subarcs. Let I= {t: a :,; t:,; b} be decomposed into the subdivision Ik = {t: tk_ 1 :,; t:,; tk}, k = 1, 2, ... , n, with t 0 = a and tn = b. If C has the representation 1: I-+ V,., define fi: Ik -+ VN as the restriction of I to the interval Ik. Then each function
h
determines a directed arc c
ck and we may write = c1 + c2 + .. · + ck.
The following result is an immediate consequence of the basic properties of Riemann-Stieltjes integrals. Theorem 16.14
(a) Suppose that Cis a rectifiable arc and that C = C1 is continuous on C, then
r~g(r)·dr = Jc (b) Suppose that continuous on
C has
±Jckr~
+ C2 + ··· + Cn. If g (16.31)
g(r)·dr.
k=1
a piecewise smooth representation
ICl. Then fcg(r)·dr=
r
I
and that g is
(16.32)
g[l(t)]·f'(t)dt.
Remark. If I is piecewise smooth then the integral on the right in (16.32) may be evaluated by first decomposing C into subarcs, each of which has a smooth representation, and then using (16.3) to add up the integrals evaluated by a smooth I on the individual subarcs. We give an illustration. EXAMPLE 2. In V3, let g = 2x 1e 1 - 3x 2 e2 + x 3e 3 and define the arc Cas the union c1 + c2 where c1 is the directed line segment from (1, 0, 1) to (2, 0, 1),
and C2 is the directed line segment from (2, 0, 1) to (2, 0, 4) (see Figure 16.7). Find the value of Jc-g(r) · dr.
= {r: r = x1e1 + eJ, 1 :,; x1 :,; 2} and c2 = {r: r = 2e1 4}. On c1, we have dr = (dx 1 )e 1
Solution. c1
1 :,;
XJ :,;
Therefore
r~ g·dr = Jc,
f
2 1
2x1 dx1 = 3
and
i
~ g·dr = c2
f
4
x 3 dx 3 =
+ x3eJ,
15
2·
1
We conclude that
D
16. Vector Field Theory; the Theorems of Green and Stokes
438
(2. 0. I)
Figure 16.7 If a functi~p g is continuous on a domain Din IRN, then Jcu ·dr will depend on the path C chosen in D. However, there are certain situations in which the value of the integral will be the same for all paths Cin D which have the same endpoints. Under such circumstances we say the integral is independent of the path. The next result illustrates this fact.
Theorem 16.15. Suppose that u is a continuously differentiable scalar field on a domain D in !RN and that p, q are points of D. Let C be any piecewise smooth path with representation f such that f(t) ED for all ton I= {t: a~ t ~ b}, the domain of f. Suppose that f(a) = p, f(b) = q. Then
fc Vu·dr = u(q)- u(p). PROOF. Because of Part (a) in Theorem 16.14 it suffices to prove the result for f smooth rather than piecewise smooth. Let e 1 , e 2 , ••• , eN be an orthonormal basis. Define for t E /, G(t) = u[f(t)] with f(t)
= f 1 (t)e 1 + · ·· + fN(t)eN.
Using the Chain rule we find
N
G'(t)
=
L u,;[/(t)]/;'(t) = Vu[f(t)] · f'(t). i=l
Hence
fc Vu · dr =
r
Vu[f(t)] · f'(t) dt
=
r
G'(t) dt
= G(b)-
= u(q) - u(p).
Next we establish a converse of Theorem 16.15.
G(a)
D
439
16.3. Line Integrals in IRN
Po
Figure 16.8
Theorem 16.16. Let v be a continuous vector field with domain D in IRN and range in VN. Suppose that for every smooth arc Clying entirely in D, the value of
fc v·dr is independent of the path. Then there is a continuously differentiable scalar field u on D such that Vu(p) = v(p) for all p e D.
PRooF. Let p0 be a fixed point of D and suppose that Cis any smooth path in D from p 0 to a point p. Define u(p) =
fc v· dr,
which, because of the hypothesis on v, does not depend on C. Let p 1 be any point of D and C0 and arc from Po to p 1 . Extend C0 at p 1 by adding a straight line segment l which begins at p 1 in such a way that the extended arc C0 + l is smooth. Denote by a the unit vector in the direction l (see Figure 16.8). We introduce a coordinate system in IRN and designate the coordinates of p 1 by x 0 • Then any point q on l will have coordinates x 0 + ta forte IR 1 . Thus, if h > 0 we find
Therefore lim -h1 [u(x 0
+ ha) -
u(x 0 )] = v(x 0 ) ·a
+ ha) -
u(x 0 )] = -lim _kl [u(x 0
h-+O+
lim _hl [u(x 0 h~o-
-
ka) - u(x 0 )]
k~o
=
-v(x 0 )·( -a)= v(x 0 )·a.
440
16. Vector Field Theory; the Theorems of Green and Stokes
This procedure can be carried out for any arc C0 passing through p 1 with a a unit vector in any direction. From Theorem 16.7 it follows that Vu =vat x 0 , D an arbitrary point of D. Suppose that u is a smooth scalar field in IR 3 • We saw earlier that if v = Vu then it follows that curl v = 0. On the other hand if v is a given vector field such that curl v = 0 in a domain D, it is not necessarily true that v is the gradient of a scalar field in D. In fact, the following example shows that the integral of v may not be path-independent, in which case v cannot be the gradient of a scalar field. To see this, we set v =(xi+ xn- 1 ( -Xze1 + x1e2 + O·e3)
with D = {(x 1 , x 2 , x 3 ): t ~xi+ x~ ~ t, -1 ~ x 3 ~ 1}. We choose for C the path r(t) =(cos t)e 1 +(sin t) · e 2 + 0 · e 3 , 0 ~ t ~ 2n. Then
fc. v·dr =I:" dt = 2n. However, if the integral were independent of the path its value should be zero since the initial and terminal points of Care the same. The difficulty arises because the path C encloses a singularity of von the line x 1 = x 2 = 0 whereas the cylindrical domain D does not contain this line. If a smooth vector field v is defined in a domain which is not merely connected but also simply connected (as defined below), then we shall prove that if curl v = 0 then v is the gradient of a scalar field u.
Definition. Let D be a domain in !RN. Then Dis simply connected if and only if whenever / 0 and / 1 are paths from I= {t: a~ t ~ b} into D with / 0 (a) = / 1 (a) and / 0 (b) = / 1 (b), there exists a function f(t, s) which is continuous for t e I and s e J = {s: 0 ~ s ~ 1}, has range in D, and has the properties: f(t, 0) = fo(t), f(t, 1)
= /1(t),
/1 (a)
for all
s E J,
f(b, s) = /o(b) = / 1 (b)
for all
s E J.
f(a, s) = fo(a) =
(16.33)
In other words, a domain D is simply connected if any two paths situated in D which have the same starting and ending points can be deformed continuously one into the other without leaving D (see Figure 16.9). The following technical lemma shows that in general the function f(t, s) for deforming one path into another in a simply connected domain can be chosen so that it has certain smoothness properties. For the proof, see Problem 17 at the end of this section and the hints given there.
u~~d
441
16.3. Line Integrals in IRN
not simply connected
simply connected
Figure 16.9
Lemma 16.4. Let I= {t: a~ t ~ b} and J = {s: 0 ~ s ~ 1} be intervals. Let D be a simply connected domain in IRN and suppose that lo and 11 with domain I are smooth paths in D for which l 0 (a) = 11 (a) and l 0 (b) = 11 (b). Then there is a continuous function l(t, s) with domain R = I x J and with range in D satisfying conditions (16.33) and also the conditions: (i) (ofot) I is continuous in R; and (ii), there is a sifficiently large integer n so that, for each t e I, the function 1 is linear ins on J; = {s: (1/n)(i- 1) ~ s ~ (1/n)i}, i = 1, 2, ... , n, with oflos and o 21/otos uniformly continuous on each rectangle I x J;.
Theorem 16.17. Suppose that vis a continuously differentiable vector field on a simply connected domain D in IR 3 and that curl v = 0 on D. Then there is a continuously differentiable scalar field u on D such that v = Vu. PROOF. We need only show that Jc:v·dris independent of the path. To do this let p and q be two points of D and let lo and 11 represent two smooth paths Co and c1 in D such that lo(a) = 11 (a) = p and lo(b) = 11 (b) = q. We choose a coordinate system with orthonormal basis e 1, e2 , e3 and write v = v1e1 + v2e 2 + v3 e 3 • From Lemma 16.4 there is a function l(t, s) = f 1 e 1 + f 2e 2 + f 3 e3 which is differentiable in t, piecewise linear ins and satisfies conditions (16.33). We define the function
q>(s) =
Lb [v1(1)(ofdot) + v2(1)(of2/ot) + v3(1)(of3/ot)] dt.
Then from (16.33) it follows that q>(O) =
fb v(l0)·I~ dt = l ~ a
q>(1)=Jb v(ld·l{dt= a
=
Jc
v · dr,
0
~~ v·dr. Jc,
We shall show that q>'(s) 0, which implies that the desired integral is independent of the path. Since v, f, and olfot are continuous, it follows that q> is continuous for s e J = {s: 0 ~ s ~ 1}. On each subinterval J; = {s: (1/n)(i- 1) ~ s ~ (1/n)i}, we know that olfos and o21/otos are uniformly
16. Vector Field Theory; the Theorems of Green and Stokes
442
continuous. Therefore we may differentiate cp and obtain
cp'(s) =
f .L b 3
{
a2J; 3 oJ; oJ;} !l• + ~ v;,i(f)-i- !l 1 dt, usut J=l ut uS
v;(/) !l
a •=1
i- 1
i
n
n
--<s 0 such that the ball with center at f*(t, s) and radius p is in D for all (t, s) e R. (b) Define a sequence {g.;}, i = 0, 1, 2, ... , n, n = 1, 2, ... , such that g. 0 (t) = f*(t, s) = f 0 (t), g •• (t) = f*(t, 1) = / 1 (t), and such that, for 1 ~ i ~ n - 1, we set 9:; equal to a polynomial with the property that l9:;(t) - f*(t, i/n)l < p/8 for a ~ t ~ b. Then let 9.;(t) = 9:;(t) - l.;(t)
where l.;(t) is the linear function coinciding withg:;(t)- f*(t, i/n) fort= a and t = b. Show that ll.;(t)l < p/8 on a ~ t ~ b and that l9.;(t) - f*(t, i/n)l
S2 , ... , S,. and their centers by P;, i = 1, 2, ... , n. According to the process described in Section 15.4, we can find mollifiers 1/1 1 , 1/12 , ••• , 1/1,. of class C'"' on an open set G0 ::::> i5 such that each 1/1; vanishes on G0 - F; where F; is a compact subset of the open rectangle S; containing P;. For each point P define
:Lr=
Then each cp; is of class coo and, for every P,. e G0 , it follows that 1 cp;(P) = 1. The functions cp 1 , cp2 , ... , cp,. are called a partition of unity (see Problem 2 of Section 15.4). We define V;
= lp;V,
i
= 1, 2, ... , n.
16. Vector Field Theory; the Theorems of Green and Stokes
452
c,
[]··
G
Figure 16.16. A finite number of rectangles covers
i5.
Then each V; is continuously differentiable on G0 and v = v1
+ v 2 + · · · + vn.
Hence it suffices to establish (16.40) for each v;. If S; is contained in D, we have
I-
aD
v;·dr = 0
since v; is zero at every point of iJD. Also, by Lemma 16.5
f
Js,r curl V; dA =I-.as, V;·dr = 0,
curl v;·dA =
D
--+
since v; = 0 except in F;, and iJS; is disjoint from F;. Finally, suppose S; is a -+ QQ.und~y rectangle. Let D; = D n S; and C; = iJD n S;. Then, since v; = 0 on iJD- C;, and on D- D;, it follows from Lemma 16.5 that
I -
m
= v.·dr J.
i =I .....
~
v.·dr l
--+
~
v.·dr l
=I
~
curl v. dA l
=I
curl. dA. l
D
Performing this process for each i and adding the results, we obtain (16.40). D If we set v = P(x 1 , x 2 )e 1 + Q(x 1 , x 2 )e 2 and express (16.40) in terms of coordinates we have the following form of Green's theorem.
16.4. Green's Theorem in the Plane
453
Corollary. If P and Q are smooth in a domain G in IR 2 which contains a regular region D, then
I ( ~Q-~P)dA=rh D
uXl
uX2
Jon
(Pdx 1 +Qdx 2)
(16.41)
where the line integral is taken in a counterclockwise direction. EXAMPLE
1. Given the disk K = {(x 1 , x 2): xi+ x~ < 1}, use Green's theorem
to evaluate
Solution. Setting P
= 2x 1 -
rh (P dx 1 + Q dx 2)
JaK
x~, Q
= x~ + 3x~, we see from (16.41) that
=IK3(xi + x~)
dA = 3
fJ 2 "
0
1
0
r 2 ·r dr d(} =
~n. D
EXAMPLE 2. Let K = {(x 1 , x 2 ): xi + x~ ~ 1} be the unit disk and let D be the region outside K which is bounded on the left by the parabola x~ = 2(x 1 + 2) and on the right by the line x 1 = 2 (see Figure 16.17). Use Green's theorem to evaluate
Ic, ( where
2
xl
x2
2
+ x2
dxl
+
2
Xt
C1 is the outer boundary of D.
Figure 16.17
xl
2
+ x2
dx2)
16. Vector Field Theory; the Theorems of Green and Stokes
454
xn
and note that Solution. We write P = -x 2 /(xi + xn, Q = xtf(xi + -"-+ - P, 2 = 0 in D. Hence with oK oriented in the counterclockwise sense, Green's theorem implies that
Q, 1
0=
I-
(P dx 1
+ Q dx 2 ) =
f- (P dx
+ Q dx 2 ) -
1
I-
Using the representation x 1 =cos(), x 2 =sin(), -n integral on the right, we obtain
fc,
(P dx 1
+ Q dx 2 )
=
I"
_,
(P dx 1
+ Q dx 2 ).
ilK
C1
iJG
(sin 2
()
~
()
~
+ cos 2 ()) d() =
n, for the last
D
2n.
EXAMPLE 3. Let D be a regular region with area A. Let v = -!(x 2 e 1 Show that A=
I-
-
x 1 e 2 ).
v·dr.
iJD
Solution. We apply Green's theorem and find that
I- I v · dr =
curl v dA =
D
iJD
I (! D
2
+
!) 2
D
dA = A.
Example 3 shows that the area of any regular region may be expressed as an integral over the boundary of that region.
PROBLEMS In each of Problems 1 through 8 verify Green's theorem. 1. P(x 1 , x 2 ) = -x 2 , Q(x 1 , x 2 ) = x 1 ; D = {(x 1 , x 2 ): 0 ~ x 1 ~ 1, 0 ~ x 2 ~ 1}.
2. P(x 1 , x 2 ) = x 1 x 2 , Q(x 1 , x 2 ) = -2x 1 x 2 ; D = {(x 1 , x 2 ,: 1 ~ x 1 ~ 2, 0 ~ x 2 ~ 3}. 3. P(x 1 , x 2 ) = 2x 1 x 2 ~ 1}.
-
3x 2 , Q(x 1 , x 2 ) = 3x 1 + 2x 2 , D = {(x 1 , x 2 ): 0
~
x1
~
2, 0
~
4. P(x 1 , x 2 ) = 2x 1 - x 2 , Q(x 1 , x 2 ) = x 1 + 2x 2 ; Dis the region outside the unit disk, above the curve x 2 = xi - 2, and below the line x 2 = 2. 5. P =xi- x~, Q = 2x 1 x 2 ; Dis the triangle with vertices at (0, 0), (2, 0), and (1, 1). 6. P = -x 2 , Q = 0; Dis the region inside the circle xi+ x~ = 4 and outside the circles xi + (x 2 - 1) 2 = 1/4 and xi + (x 2 + 1) 2 = 1/4. 7. v =(xi+ x~r 1 (-x 2 e 1 + x 1 e2 ); D = {(x 1 , x 2 ): 1 <xi+ x~ < 4}. 8. P = 4x 1 - 2x 2 , Q = 2x 1 + 6x 2 ; Dis the interior ofthe ellipse: x 1 = 2 cos 8, x 2 = sin 8, - n ~ 8 ~ n. 9. Prove Part (b) of Theorem 16.19.
455
16.5. Surfaces in IR 3 ; Parametric Representation 10. Given D = {(x 1 , x 2): 0::::;; xii3 region.
+ x~l3
t 1 ) =F (s 2 , t 2 ). In other words, Equations (16.42) define a one-to-one C 1 transformation from D u oD onto the points of the surface element. Observe that if r(s, t) = x 1 (s, t)e 1
+ x 2 (s, t)e 2 + x 3 (s, t)e3 ,
then
r.
X
- (ox2 OXJ - ox2 ox3) 1 (OXJ oxl - OXJ oxl) 2 OS ot ot OS e + OS ot ot OS e + (oxl ox3- oxl OXJ)e3. os at at os
r, -
16.5. Surfaces in IR 3 ; Parametric Representation
457
Employing the Jacobian notation which we introduced on page 359
J(~~)=
0(()
0(()
;;
ov = (ocp oi/J _ ocp oi/J) oi/J ou ov ov ou • ov
ou
(16.45) We say that the equations (16.42) or (16.43) form a parametric representation of the smooth surface element. The quantities s and t are called parameters. The next result shows the relation between two parametric representations of the same surface element.
Theorem 16.21. Suppose that the transformation given by (16.43) satisfies Condition (16.44) and that (16.43) is C 1 on a region G which contains D u oD. Let S and S1 denote the images of D u oD and G, respectively. (a) If (s, t) is any point of D u oD, then there is a positive number p such that
the part of S corresponding to the disk (s -
si + (t -
t)2 < P2
has one of the forms X3 = f(xl, X2),
where f, g, or his smooth near the point (x1 , x2 , x3 ) co"esponding to (S, t). (b) Suppose that another parametric representation of S and S1 is given by
-
v(OP) = r 1 (s', t'),
(s', t') e D1 u oD 1 ,
in which S and S1 are the images of D1 u oD1 and G1 respectively, in the (s', t') plane. Assume that D1 , G1 and r 1 have all the properties which D, G, and r have. Then there is a one-to-one transformation T: s = U(s', t'),
t = V(s', t'),
(s', t') e G1 ,
(16.46)
form G1 to G such that T(Dd = D, T(oDd = oD and r[U(s', t'), V(s', t')] = r 1 (s', t') for (s', t') e G1 • PROOF
(a) Since '• x r, :F 0 at a point (S, i), it follows that at least one of the three Jacobians in (16.45) is not zero at (S, t). Suppose, for instance, that
J(x1,s, tx2) #: 0
16. Vector Field Theory; the Theorems of Green and Stokes
458
at (S, t). We set r(s, t) = X 1(s, t)e 1 + X 2 (s, t)e 2
+ X 3(s, t)e 3
and denote x1 = X 1(S, i), x2 = X 2 (S, i), x3 = X 3(S, t). Then from the Implicit function theorem it follows that there are positive numbers IX and {J such that all numbers x 1, x 2 , s, t which satisfy (x 1
-
x- 1 )2 + (x 2
-
x- 2 )2 < IX 2,
and lie on the graph of s = q>(xt> x 2),
t
= 1/J(xl, x2),
where q> and 1/1 are smooth functions in the disk (x- xd 2 + (x 2 In this case the part of S1 near (xt> x 2 , x3) is the graph of x3 = X3[q>(x1, x2), ljl(x1, x 2)]
for (x 1 -
-
x2)2 < 1X2.
x1)2 + (x 2 - x2)2 < IX2.
The conclusion (a) follows when we select p > 0 so small that the image ofthe disk(s- sjl + (t- f} 2 < p 2 liesinsidethedisk(X1- Xd 2 + (x2- X2) 2 < IX 2. (b) Since r(s, t) and r 1(s', t') are one-to-one, it follows that to each (s', t') in G1 there corresponds a unique P on S1 which comes from a unique pair (s, t) in G. The relationship is shown in Figure 16.19. The transformation T in (16.46) is defined by this correspondence: s = U(s', t'), t = V(s', t'). Then Tis clearly one-to-one. To see that Tis smooth, let (s0, t 0) be any point in G1 and let s0 = U(s 0, t0), t 0 = V(s 0, t 0). Denote by P0 the point of S corresponding to both (s 0 , t 0 ) and (s0, t 0). At least one of the three Jacobians in (16.45) does not vanish at (s 0 , t 0 ). If, for example, the last one does not vanish, we can solve for sand tin terms of x 1 and x 2 as in Part (a). We now set r 1(s', t')
= XJ.(s', t')e 1 + X2(s', t')e2 + X3(s', t')e 3 •
At points P of S near P0 the surface can be represented as a function of (x 1 , x 2) so that there is a one-to-one correspondence between the range and domain
0
Figure 16.19. Defining the transformation T.
459
16.5. Surfaces in IR 3 ; Parametric Representation
of this function. Therefore we find U(s', t') =
dA"'6
7t/4
0
0
"/4
0
(1
(sin 2 q> sin 2
+ cos 2
(}
+ cos 2 q>) sin q> dq> =
cp) sin q> d(} dq> nJ 12 (16 - 7 j2).
0
Remark. The parametric representation of S in spherical coordinates does not fulfill the required conditions for such representations since lr"' x r6 1= sin q> vanishes for q> = 0. However, the surface S with the deletion of a small hole around the x 3 axis is of the required type. We then let the size of the hole tend to zero and obtain the above result for lx 1 • ExAMPLE 3. Given R = { (x 1 , x 2 , x 3 ): xi + x~ ::;; 1, 0 ::;; x 3 ::;; x 1 + 2} and S = oR. If S has uniform density J, find its mass. Find the value of
and obtain x 1 , the xrcoordinate of the center of mass of S (see Figure 16.26). Solution. The surface is composed of three parts: S1 , the disk in the x 1 x 2 -plane; S2 , the lateral surface of the cylinder; and S3 , the part of the plane x 3 = x 1 + 2
(I. 0. 3)
Ixi+ x~ = 1 x2 I
( -I. 0, I)
1
I I
V
s2
/
I
---+-7"-
--*-0
s,
Figure 16.26
I
16.6. Area of a Surface in IR 3 ; Surface Integrals
469
inside the cylinder. We have (since x 1 is an odd function)
On S3 we see that x 3 = x 1 S1 • Hence
+ 2 so that dS = J2 dAis the element of area on
On S2 we choose coordinates (0, z) with x 1 dAez· We write
0 ~ z ~ 2 + cos 0}
D2 = {(0, z): -n: ~ 0 ~ n,
and find
fi
dS =
X1
S2
If
cos 0 dA 6z
=
= cos 0, x 2 = sin 0, x 3 = z, dS =
f"
f2+cos9
-x
D2
COS
0 dz dO
= n.
0
Therefore
The surfaceS is a piecewise smooth surface without boundary. The intersections of x~ + x~ = 1 with the planes x 3 = 0, x 3 = x 1 + 2 form two smooth edges. There are no vertices on S. To find the mass M(S) of S, we observe that
=
M(S) Clearly, A(Sd =
fI
{J dS = {JA(S) = {J[A(Sl)
+ A(S2) + A(S3)].
n, A(S3) = n.J2. Also,
A(S2) =
fl
dA 6 z =
f"
j2+•ose dz dO= 4n.
JD Jo Therefore M(S) = n:{J(5 + J2). To obtain the x 2
-x
1
coordinate of the center of
mass, note that
Hs x
_ {J {Jn: 1 dA 1 x = M(S) = {Jn:(5 +
J2)
1
5+J2"
0
PROBLEMS
In each of Problems 1 through 9, find the value of
fI
1. f(x 1 , x 2 , x 3 )
f(x 1 , x 2 , x 3 ) dS.
= x 1 , S = {(x 1 , x 2 , x 3 ): x 1 + x 2 + x 3 = 1, x 1 ;;:?; 0, x 2 ;;:?; 0, x 3 ;;:?; 0}.
16. Vector Field Theory; the Theorems of Green and Stokes
470
2. f(x 1 , x 2 , x 3) =xi, S = {(x 1 , x 2 , x 3): x 3 = x 1 , xi+ x~:::;; 1}. 3. f(x 1 , x 2 , x 3) =xi, S = {(x 1 , x 2 , x 3): x~ =xi+ x~, 1:::;; x 3 :::;; 2}.
4. f(x 1 , x 2 , x 3) =xi, Sis the part of the cylinder x 3 = xi/2 cut out by the planes x 2 = 0, x 1 = 2, and x 1 = x 2 . 5. f(x 1 , x 2 , x 3) = x 1 x 3, S = {(x 1 , x 2 , x 3): xi+ x~ = 1, 0:::;; x 3 :::;; x 1 + 2}. 6. f(x 1 , x 2 , x 3) = xl> Sis the part of the cylinder xi - 2x 1 + x~ = 0 between the two nappes of the cone xi+ x~ = x~. 7. f(x 1 , x 2 , x 3) = 1; using polar coordinates (r, 8) in the x 1 , x 2 -plane, Sis the part of the vertical cylinder erected on the spiral r = 8, 0 :::;; 8 :::;; n/2, bounded below by the x 1 xz-plane and above by the cone xi+ x~ = x~. 8. f(xl, x 2 , x 3) =xi+ x~- 2x~, S = {(x 1 , x 2 , x 3): xi+ x~ + x~ = a 2 }. Xz, x3) =xi, s =oR where R = {(xl> Xz, x3): X~~ xi+ X~, 1:::;; x3:::;; 2} (see Problem 3).
9. f(xl,
In each of Problems 10 through 14, find the moment of inertia of S about the indicated axis, assuming that the density (J is constant. 10. The surfaceS of Problem 3; x 1 axis. 11. The surface S of Problem 6; x 1 axis. 12. The surface S of Problem 7; x 3 axis. 13. The surfaceS which is the boundary of R, where R = {(x 1 , x 2 , x 3): x 1 + x 2 +
x 3 < 1, x 1 > 0, x 2 > 0, x 3 > 0}; about the x 2 axis. 14. The torus S = {(x 1 , x 2 , x 3): (Jxi + x~- b) 2 + x~ = a2 , 0 are introduced by the relations x 1 = (b +a cos cp) cos 8 x 2 = (b + a cos cp) sin 8 x 3 =a sin q>,
the torus Sis described by {(cp, 8): 0:::;; cp:::;; 2n, 0:::;; 8:::;; 2n}.]
In each of Problems 15 through 17 find the center of mass assuming the density (J is constant. 15. Sis the surface of Problem 2. 16. S is the surface of Problem 13.
The electrostatic potential E(Q) at a point Q due to a distribution of electric charge (with charge density p) on a surfaceS is given by E(Q) =
fLp(:~QdS
where dPQ is the distance from a point Q in IR 3
-
S to a point P
E
S.
16.7. Orientable Surfaces
471
n
Figure 16.27. The unit normal to the surfaceS.
In Problems 18 through 21 find E(Q) at the point given, assuming that p is constant. 18. S = {(x 1 , x 2 , x 3 ): xi+ x~ = 1, 0 ~ x 3 ~ 1}; Q = (0, 0, 0). 19. S = {(x 1 , x 2 , x 3 ): xi+ x~ + x~ = a 2 }; Q = (0, 0, c). Case 1: c >a> 0; Case 2: a>c>O. 20. S = {(x 1 , x 2 , x 3 ): xi+ x~ + x~ = a2 , x 3 ~ 0}. Q = (0, 0, c); 0 < c 0.
3. v = xie 1 + x~e 2 + x~e 3 ;
-
1)2 + x~.::; 1},
{(x 1 , x 2, x 3): xi+ x~ = x~; 1.::; x 3 .::; 2}, n·e 3 > 0.
S=
4. v = x 1x 2e 1 + x 1x 3e2 + x 2x 3e3; S = {(x 1 , x 2, x 3): x~ = 2- x 1 ; x~.::; x 2 .::; x~12 }. 5. v = x~e 1 + x 3e 2 - x 1e 3; n·e 1 > 0. 6. v = 2x 1 e 1 n·e 1 > 0.
-
S=
x 2e 2 + 3x 3e 3;
{(x 1 , x 2, x 3): x~ = 1- x 1 , 0.::; x 3 .::; x 1 ; x 1 ~ 0},
S=
{(x 1 , x 2, x 3): x~ = x 1 ; x~.::; 1- x 1 ; x 2 ~ 0},
In each of Problems 7 through 12, verify the Stokes theorem. 7. v = x 3e 1 + x 1e 2 + x 2e 3; S = {(x 1 , x 2, x 3): x 3 = 1- xi- x~, x 3 ~ 0}, n·e 3 > 0. 8. v = x~e 1 + x 1x 2e2 - 2x 1x 3e 3; n·e 3 >0. 9. v = -x 2x 3e 3; the sphere.
S=
{(x 1 , x 2, x 3): xi+ x~ + x~ = 1,
x 3 ~ 0},
x~ = 4, xi+ x~ ~ 1}, n pointing outward from
S ={(xi+ x~ +
10. v = -x 3 e 2 + x 2 e 3 ; Sis the surface of the cylinder given in cylindrical coordinates by r = 0, 0 .::; (J .::; n/2, which is bounded below by the plane x 3 = 0 and above by the surface of the cone xi + x~ = x~; n · e 1 > 0 for (J > 0.
11. V = X2e1
+ X3e2 + X1e3; S = {(xl, X2, X3): X~= 4- Xl, X1 ~ xn, n•el > 0.
12. v = x 3 e 1 - x 1 e 3 ; Sis the surface of the cylinder given in cylindrical coordinates by r = 2 +cos (J above the plane x 3 = 0 and exterior to the cone x~ =xi+ x~; n is pointing outward from the cylindrical surface.
In each of Problems 13 and through 15 use the Stokes theorem to compute
Jasv ·dr. 13. v = 2.
r- 3r where r =
x 1 e1 + x 2e 2 + x 3e 3 and r =
lrl; Sis the surface S2 of Example
14. v =(ex' sin x 2 )e 1 +(ex' cos x 2 - x 3)e2 + x 2e 3; Sis the surface in Problem 3.
s
= {(xl, x2, x3): xi +X~+ X~= 15. v =(xi+ x3)el + (xl + xDe2 + (x2 + xne3; 1, x 3 ~ (xi + xD 112}; n points outward from the spherical surface. 16. Show that if Sis given by x 3 = f(x 1, x 2 ) for (x 1 , x 2 ) ED= {(x 1 , x 2 ): xi+ x~.::; 1}, iff is smooth, and if v = (1- xi- xDw(x 1, x 2, x 3) where w is any smooth vector field defined on an open set containing S, then
f
fs(cur! v)·n dS = 0.
17. Suppose that v = r- 3(x 2e 1 + x 3e 2 + x 1 e 3) where r = x 1 e 1 + x 2e 2 + x 3e 3 and r = lrl, and Sis the sphere {(x 1 , x 2, x 3): xi + x~ + x~ = 1} with n pointing outward. Show that
I
f
(curl v)·n dS = 0.
16. Vector Field Theory; the Theorems of Green and Stokes
486
18. Suppose that a smooth surfaceS has two different smooth parametric representations, r(s, t) for (s, t) E D and r 1(s', t') for (s', t') E D1 . Let v be a smooth vector field defined on S. Find the relationship between the formulas for (curl v) · n in the two representations. 19. In the proof of the Stokes theorem, carry out the verification that all the second derivative terms cancel in the displayed formula before (16.68). 20. Let M be a Mobius strip. Where does the proof of the Stokes theorem break down for this surface?
16.9. The Divergence Theorem Green's theorem establishes a relation between the line integral of a function over the boundary of a plane region and the double integral of the derivative of the same function over the region itself. Stokes's theorem extends this result to two-dimensional surfaces in three-space. In this section we establish another kind of generalization of the Fundamental theorem of calculus known as the Divergence theorem. This theorem determines the relationship between an integral of the derivative of a function over a three-dimensional region in IR 3 and the integral of the function itself over the boundary of that region. All three theorems (Geen, Stokes, Divergence) are special cases of a general formula which connects an integral over a set of points in IRN with another integral over the boundary of that set points. The integrand in the first integral is a certain derivative of the integrand in the boundary integral. Let v = v1 (x 1 , x 2 , x 3 )e 1 + v2 (x 1 , x 2 , x 3 )e 2 + v3 (x 1 , x 2 , x 3 )e 3 be a vector field defined for (x 1 , x 2 , x 3 ) in a region E in IR 3 . We recall that div vis a scalar field given in coordinates by the formula
8v 2 8v 3 8v 1 d l. V V =-+-+-. 8x 1 8x 2 8x 3 The Divergence theorem consists of proving the formula (16.73) where oE is oriented by choosing n as the exterior normal to 8E. We first establish (16.73) in several special cases and then show that the formula holds generally, provided that the boundary of E is not too irregular and that vis smooth. Lemma 16.7. Let D be a domain in the (x 1 , x 2 )-plane with smooth boundary. Let f: D u 8D-+ IR 1 be a piecewise smooth function and define E
= {(x 1 , x 2 , x 3 ): (x 1 , x 2 ) ED, c < x 3 < f(x 1 , x 2 )}
487
16.9. The Divergence Theorem
XJ
Figure 16.37. aE
S1 v S2 v S3 .
=
for some constant c. Suppose that v = u(x 1 , x 2 , x 3 )e3 is such that u and au;ax 3 are continuous on an open set in ~ 3 containing E u aE. Then
where n is the outward unit normal of aE. PROOF. Since div v = aujax 3 , it follows that
IIL
div
V
dV =
IIL::
3
dV =
IL1/(x~ox,) ::
3
dx 3 dAx,x,·
Performing the integration with respect to x 3 , we find
IIL
div v dV
=I L
{u[xl> x 2 , f(xl> x 2 ) ] - u(x 1 , x 2 , c)} dAx,x,·
(16.74)
Let aE = S1 u S2 u S3 where S1 is the domain in the plane x 3 = c which is congruent to D; S2 is the lateral cylindrical surface of aE; and S3 is the part of aE corresponding to x 3 = f(x 1 , x 2 ) (see Figure 16.37). We wish to show that (16.74) is equal to
If
s,us,us3
v·n dS.
Along S2 , the unit normal n is parallel to the (x 1 , x 2 )-plane, and son· e 3 = 0. Therefore v · n = 0 on S2 and
II, II. -IL
v · n dS = 0.
(16.75)
The outer normal along S 1 is clearly -e3 and therefore
v·n dS =
u(x 1 , x 2 , c) dAx,x,·
(16.76)
488
16. Vector Field Theory; the Theorems of Green and Stokes
As for S3 , the unit normal function is given by
of ) 2 dS = [ 1 + ( oxl
+
(of )2]112 ox2 dAx,x,·
Therefore, since v = ue 3 , we find
fL
3
v·ndS=
fL
u[x 1 ,x2,f(x1 ,x2)]dAx,x,·
(16.77)
The result follows when (16.74) is compared with (16.75), (16.76), and (16.77).
D Lemma 16.8. Suppose that the hypotheses of Lemma 16.7 hold, except that v has the form
v = v1 (x 1 , x 2, x 3 )e 1
+ u2 (x 1 , x 2, x 3)e2
with u 1 , u 2 smooth function on an open set containing E u oE. Then
where n is the outward unit normal of oE. PROOF.
Let U1 (x 1 , x 2, x 3 ), U2(x 1 , x 2, x 3 ) be defined by
U1 (x 1 , x 2, x 3 ) =
-1"
3
u 1 (x 1 , x 2, t) dt, U2 (x 1 , x 2, x 3 ) =
1"
3
u2 (x 1 , x 2, t) dt.
In addition, we define
Then w is a smooth vector field and U3 , oU3jox 3 are continuous, and hence curl w = - oul - e1 ox3 . dtvv
oul oxl
oU2 +e2 + (oul - -oU2) - e3 = v + u,
ou2 ox2
ox3
oU3 ox3
oxl .
= - + - = - = -d1vu.
ox2
(16.78)
Since oE is a piecewise smooth surface without boundary, the Corollary to
489
16.9. The Divergence Theorem
Stokes's theorem is applicable. Hence
fIE
(curl w)·n dS = 0
=fIE
(v
+ u)·n dS.
Therefore using Lemma 16.7 for the function u = - U3 e 3 , we find
fIE
-fIE
ffL
div u dV.
(16.79)
The result of the lemma follows by inserting (16.78) into (16.79).
D
v • n dS =
u • n dS = -
The Divergence theorem will now be established for a wide class of regions which are called "regular". Intuitively, a region in IR 3 is regular if its boundary can be subdivided into small pieces in such a way that each piece has a piecewise smooth representation of the form x 3 = f(x 1 , x 2 ) if a suitable Cartesian coordinate system is introduced. Definition. A region E in IR 3 is regular if and only if: (i) oE consists of a finite number of piecewise smooth surfaces, each without boundary; (ii) at each point P of oE a Cartesian coordinate system is introduced with Pas origin. There is a cylindrical domain r = {(x 1 , x 2 , x 3 ): (x 1 , x 2 ) e D, -oo < x 3 < oo }, with D a region in the plane x 3 = 0 containing the origin, which has the property that r n oE is a surface which can be represented in the form x 3 = f(x 1 , x 2 ) for (x 1 , x 2 ) e D u oD. Furthermore, f is piecewise smooth (see Figure 16.38); (iii) the set
r1 =
{(x 1 , x 2 , x 3 ): (x 1 , x 2 ) e D, -c < x 3 < f(x 1 , x 2 )}
for some positive constant c (depending on P) is contained entirely in E. Remarks. If E is a regular region, then each point P of oE is interior to a smooth surface element except for those points on a finite number of arcs on oE which have zero surface area. If a line in the direction of the unit normal
Figure 16.38. A regular region.
490
16. Vector Field Theory; the Theorems of Green and Stokes
to 8E is drawn through a point P interior to a smooth surface element on 8E, then a segment of the line on one side of P will be in E, while a segment on the other side will be exterior to E. Finally, we observe that n varies continuously when Pis in a smooth surface element of 8E. Theorem 16.23 (The Divergence theorem). Suppose that E is a closed, bounded, regular region in IR 3 and that v is a continuously differentiable vector field on an open set G containing E u 8E. Then
where n is the unit normal function pointing outward from E. PROOF. With each point P E 8E associate a coordinate system and a cylindrical domain r as in the definition of a regular region. Let rP be the bounded portion of r such that - c < x 3 < c (where cis the constant in the definition of regular region; c depends on P). With each interior point P of E, introduce a Cartesian coordinate system and a cube rP with Pas origin, with the sides of rP parallel to the axes, and with fP entirely in E (see Figure 16.39). Since E u 8E is compact, a finite number r 1 , r 2 , ... , rn cover E u 8E. As described in the proof of Green's theorem, there is a partition of unity cp 1 , cp 2 , •.• , CfJn of class C'" on an open set G containing E u 8E such that each CfJ; vanishes on G- F; where F; is a compact subset of r;. We define V;
=
(/J;V,
i = 1, 2, ... , n.
x,
)-., iJE
x,
--- - -
--
--- ------
Figure 16.39
16.9. The Divergence Theorem
491
Then each v; is continuously differentiable on G and vanishes on G - F;. Also, v= 1 V;. Therefore it is sufficient to establish the result for each V;. Suppose first that P; is an interior point of E. Then f; c: E and V; vanishes on or;. Hence
Li'=
IIL
div v; dV
=
IIL.
because of Lemma 16.8. Also, because
ILE
=0
= 0 on oE, we have
V;
V; • n
div v; dV
dS = 0.
Thus the result is established for all such ri. Suppose now that P; E oE. Define E; = r; n E. Then V; = 0 on the three sets: (i) E - E;; (ii) oE - Y; where Y; = oE n r;; (iii) oE; - Y;· Now, since E; is a region of the type described in Lemmas 16.7 and 16.8, we find
IIL
div
V;
dV =
=
IIL.
It
div
V; • n
V;
dS
dV =
=
ILE,
ILE
V; • n
V; • n
dS
dS.
The result is established for all v; and hence for v. EXAMPLE
D
1. Let E be the region given by
E = {(x 1, x 2, x 3): 1 ~xi+ x~ + x~ ~ 9}. Letr = x 1 e 1 theorem. .
+ x 2 e 2 + x 3 e 3 and v = r- 3 rwhere r = ...
2
2
lrl. Verify the Divergence
2
...
•
= {(xl, x2, x3): xl + x2 + x3 = 1} and s2 = {(xl, x2, x3).
Solutwn. Let sl
xi+ x~ + x~ = 9} be the boundary spheres of E with n pointing outward from the origin 0. Then
IIai
v • n dS =
IIs,
v · n dS -
IIs,
v · n dS.
A cop.putation shows that div v = 0 and so J J JEdiv v dV = 0. On both S1 and S2 the normal n is in the radial direction and hence n = r- 1 r. Therefore
IIs
2
v·ndS-
IIs, v·ndS=~A(S2 )-A(S1 )=0.
0
EXAMPLE 2. Let E = {(x 1, x 2 , x 3): xi+ x~ < 1, 0 < x 3 < x 1 + 2} and define v = t(xi + xne 1 + t(x~ + x 3xi)e 2 + t(x~ + xix 2)e 3. Use the Divergence theorem to evaluate
16. Vector Field Theory; the Theorems of Green and Stokes
492
Solution. A computation shows that div v = x 1 disk by F and obtain
ffoE
v · n dS =
=
ffL + 2+ (x 1
x
+ x 2 + x 3 • We denote the unit
x 3 ) dV
1 (! ( 2 " 19 2 Jo Jo (3r 2 cos 2 0 + 4])r dr dO= 8 n.
D
PROBLEMS
In each of Problems 1 through 10 verify the Divergence theorem by computing div v dV and aE v · n dS separately.
JJJE
JJ
1. v = x 1x 2e 1 + x 2 x 3 e2 + x 3x 1e 3; E is the tetrahedron with vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1). 2. v = xie 1
-
x~e 2 + x~e 3 ; E = {(x 1, x 2, x 3): xi+ x~
3. v = 2x 1e 1 + 3x 2e 2 - 4x 3e 3; E
= {(x 1, x 2, x 3): xf +
< 4, 0 < x 3 < 2}. x~ + x5
< 4}.
4. v = xfe 1 + x~e 2 + x5e 3; E = {(x 1, x 2, x 3): x~ < 2- x 1, 0 < x 3 < xt}. 5. v = x 1e 1 + x 2e2 + x 3e 3; E = E 1 n E 2 where E2 = {(xl, X2, X3): Xf +X~+ X~< 4}.
E 1 = {(x 1, x 2, x 3): xf + x~ > 1},
6. v = x 1e 1 - 2x 2e 2 + 3x 3e 3; E = E 1 n E 2 where E 2 = {(x 1, x 2, x 3): x5 < 4- xt}. 7.
"= r- 3(x3el + xle2 + x2e3), r = (xf x~ + x5 < 4}.
E 1 = {(x 1, x 2, x 3): x~ < xt},
+X~+ xn 112 ; E
= {(xl, X2, x3): 1 <xi+
8. v = x 1e 1 + x 2e2 + x 3e 3; E = E 1 n E 2 where E 1 = {(x 1, x 2, x 3): xi + x~ < 4}; E2 = {(x 1 , X2, x 3 ): Xf +X~- X~> 1}.
= 2x 1e 1 + x 2e2 + x 3e 3; E = E 1 n E 2 where E 1 = {(x 1, x 2, x 3): x 3 ~ xf + xD; E 2 = {(x 1, x 2 , x 3): 2x 1 ~ x 3}.
9. v
10. v = 3x 1e 1 - 2x 2e 2 + x 3e 3; E = E 1 n E 2 n E 3 where E 1 = {(x 1, x 2, x 3): x 2 ~ 0}; E 2 = {(xto x 2, x 3): xf + x5 ~ 4}; E 3 = {(x 1, x 2, x 3): x 1 + x 2 + x 3 ~ 3}.
In each of Problems 11 through 13, use the Divergence theorem to evaluate HaEv·n dS. 11. v = x 2 e"'e 1 + (x 2 - 2x 3e"•)e 2 + (x 1e"•- x 3)e 3; E = {(xto x 2 , x 3): [(xi+ x~) 1'2 -2] 2 + x5 < 1}.
493
16.9. The Divergence Theorem
12. v
= x~e 1
+
x~e 2
+
x~e 3 ; E =
{(x 1 ,
x2,
x 3 ): xi+
x~
+
x~
< 1}.
13. v = x~e 1 + x~e 2 + x 3 e 3 ; E = {(x 1 , x 2 , x 3 ): xi+ x~ < 1, 0 < x 3 < x 1 + 2}.
14. Let E be a regular region in ~ 3 . Suppose that u is a scalar field and vis a vector field, both smooth in an open region G containing E u oE. Show that
fft
u
div v dV
=
fIE
uv • n dS -
fft
(grad u) · v dV.
15. Let E and G be as in Problem 14 and u, grad u and v smooth functions in G. Let ojon denote the normal derivative on oE in the direction of n. If ll denotes the Laplace operator: (8 2 ;a xi) + (o 2 ;ax~) + (o 2 ;axn, show that
If u is any solution of Llu = 0 in E, prove that
fIE ::
dS = 0.
Appendixes
Appendix 1. Absolute Value If a is a real number, the absolute value of a, denoted by lal, is defined by the conditions Ia I = a if a > 0, lal = -a if a< 0, 101 =0. Algebraic manipulations with absolute values are described in the following theorem. TheoremA.l
(i) (ii) (iii) (iv)
lal ~ 0, 1-al = lal, and lal 2 = a2 • Ia · bi = lal·lbl and, if b ::1= 0, then la/bl = lal/lbl. lal = ibla = ±b. If b is a positive number, then lal 1. ~
ba < b, and
< Pi-1 + 1.
Hence b 1- 1a - 1 < p1_1 or b 1a - b < bpi_1. Therefore di = Pi - bpi- 1 < bia - (b 1a - b) = b d, =Pi- bpi- 1 > (bia - 1)- b(bi-la) = -1.
Since d1 is an integer and -1 < d1 < b, we conclude that 0 ~ d1 ~ b - 1. Using the relation Pi= di + bpi_ 1 and proceeding by induction, we find that Hence, I
Pt = L bH'd,. ~ b 1a < Pi /c=1
+ 1.
But L~= 1 bi-"d,. = biL~= 1 b-"d,.. Setting si = L~= 1 b-"d,., we have s1 = b- 1P~t and so s1 ~a< si + b-i. Therefore, s1 -+ a as i-+ oo. Thus far we have shown that each number a has a proper development with the base b. To show that there is a unique proper development, let
d1 d2 d~ b' b 2 ' ••• , bn' ••• be a second proper development for the number a. That is, co d~
a=
L b~· 1=1
We may write n
b•a = L bn-1d; 1=1
+
co
L bn-td; l=n+1
Since the last sum on the right is a proper development of some number, it must be a number less than 1. Hence n
n
L bn-td; ~ b•a < L bn-idi + 1. 1=1 i=1
We define Pn = L~= 1 bn-td; and obtain Pn ~ b•a < Pn
+ 1.
Appendixes
506
Therefore these Pn are the same as the Pn used for the development ddb\ d2 /b 2 , ••• , dn/b", ... Hence d; = di for every i and the development 0 is unique. The next result is frequently useful in studying properties of the real number system.
Theorem A.6. There is a rational number between any two real numbers. PROOF. Let the given numbers be a and b with a < b. Without loss of generality assume a ~ 0, as otherwise adding a sufficiently large positive integer to a and b will make them nonnegative. Let q be the smallest positive integer larger than 1/(b- a). Let p be the smallest positive integer larger than qa. Then we have p - 1 ~ qa so that 1 p qa < p ~ qa + 1 = a < - ~ a + -. q q
Also, q(b- a)> 1 -a+ (1/q) < b, and we conclude that
a
0
q
Corollary. There is an irrational number between any two real numbers. Let the given real numbers be a and b. From Theorem A6 there is a rational number r between aj.J2 and bj.J2. Then r.J2 is between a and b 0 and r.J2 is irrational (since .J2 is-see Theorem A.7). PROOF.
We show now that .J2 is irrational.
Theorem A.7. There is no rational number whose square is 2. PROOF. Suppose there is a rational number r such that r 2 = 2. We shall reach a contradiction. If r is rational there are integers p and q such that r = pjq. Assume that p and q have no common factor, a fact which follows from the axioms of Chapter 1 (although not proved there). Since r 2 = 2, we have p2 = 2q 2 and p2 is even. The fact that p2 is even implies that pis even. For if p were odd, then p = 21 + 1 for some integer 1 and then p2 = 41 2 + 41 + 1. Thus p 2 would be odd. Hence, p = 2k for some integer k and q 2 = 2P. We see that q 2 is even and by the same argument as that used for p 2 , it follows that q is even. However, we assumed that p and q have no common factor and 0 then deduced that they have 2 as a common factor, a contradiction.
PROBLEMS
*1. Given the integers d 1 , d2 ,
say the expansion
••• ,
d" with 0
~
d;
~
b- 1 where b is an integer > 1. We
Appendix 4. Vectors in EN
507
is repeating if and only if there is a positive integer n such that d;+n = d;, i = 1, 2, ... Show that every repeating development represents a rational number. 2. Prove that
-/3 is irrational.
*3. Consider the numbers on [0, 1] represented with the base b = 3. We shall write these numbers using "decimal" notation. Describe geometrically on the interval [0, 1] all points corresponding to numbers ofthe form O.ld 2 d 3 d4 •••
where the d; may be 0, 1, or 2. Similarly, describe the sets
where the a; and b; may be 0, 1, or 2. Finally, describe the set corresponding to
where all the c; are either 0 or 2.
Appendix 4. Vectors in EN I. The space EN. We recall the definition of a vector in a plane or threespace as an equivalence class of directed line segments, all having the same length and direction. We shall extend this definition to N -dimensional space and establish several of the basic properties of such vectors. By a Euclidean N-space, denoted EN, we mean a metric space which can be mapped onto IRN by an isometry. Such a mapping is called a coordinate system and will be denoted by a symbol such as (x). If A is a point in EN and (x) is a coordinate system, then the point in IRN which corresponds to A is denoted xA or (xf, xf, ... , x:). Of course, IRN itself is a Euclidean N-space; the plane and three-space studied in elementary geometry courses are also examples of Euclidean spaces. In this section we establish several elementary properties of EN. The reader is undoubtedly familiar with the two- and three-dimensional versions of the results given here. We define a line to be a Euclidean 1-space. Ifl is a line and (t) is a coordinate system on l, then the origin and unit point of (t) are the points having coordinates 0 and 1, respectively. Theorem A.8. Suppose that l is a line and (t) and (u) are coordinate systems on l. Let the origin of the (u) system have t-coordinate t0 and the unit of the (u) system have t-coordinate t0 + A.. Then for every P on l, it follows that
uP = A.(tP - to),
A. = ± 1.
PRooF. We have luP- 01 = ltP- t0 1 and luP- 11 =It'- t 0 - A.l for all p. Squaring both sides of these equations and subtracting, we get
uP = A.(tP - to).
Appendixes
508
Choosing p and 0 as the unit point and the origin, we find
lu 1 -
U0
1= 1 = I.A.(t 1 - t0 )1 = I.A.I·It 1 - t0 1= I.A.I;
hence .A.= ±1.
D
We observe that a coordinate system (t) on a line l sets up an ordering of the points in which P1 precedes P2 if and only if t 1 < t 2 • We use the notation P1 -< P2 to indicate that P1 precedes P2 • Theorem A.S shows that there are exactly two possible orderings on every line according as .A. = 1 or -1. A line l and a specific ordering, denoted lV, determine a directed line l The same line with the 2J>POSite ordering is frequently denoted -l A pair of points (A, B) on a line l will determine a directed line segment when they are ordered with respect to lV. We use the symbol AB for such a directed line segment. If A and B are any points on a line l, we define the directed distance from A to B along ~ denoted AB, by the formula AB = AB =
IABI ifA-) f... A..(x~ I l & ' i=l
Using Theorem A.9, we find ----+
Proj-; AB = A'B' = tB'- tA' =
N
L A.;(xf- xt).
D
i=l
2. Vectors in EN. As in the case oftwo and three dimensions, we shall define a vector as an equivalence class of directed line segments. Two directed line segments AB and ---+ CD are equivalent if and only if ---+ Proj-; AB = Proj-;CD ...
----+
for every directed line l in EN. We writeAB ~ CD for such equivalent directed line segments. The proofs of the following elementary properties of equivalent directed line segments are left to the reader.
Theorem A.l7 ----+
(a) /f(x) is a coordinate system in EN, thenAB ~CD if and only if xf- xf = xf ---xt, i = 1, 2, ... , N. --+ ---+(b) If AB ~CD then CD ~AB. (c) If AB ~ CD and CD ~ EF, then AB ~ EF. (d) If AB and P are given, then there is a unique point Q such that
------
~---+
PQ ~AB. - and ---+ ~ DF. (e) If AB ~DE BC ~EF, then AC (f) If AB ~DE, if Cis h of the way from A to B, and ifF ish of the way from """""-+ D toE, thenAC ~DF.
-
Definitions. A vector in EN is the collection of all ordered pairs of points (A, B), i.e., all directed line segments AB in EN having the same magnitude and direction. The individual ordered pairs are called representatives of the vector containing them. We denote by v(AB) the vector containing the representative A B. Let v and w be vectors and AB a representative of v. Then there is a representative BC of w. We define the sum v + was the vector u which as ---+ AC
-
-
Appendix 4. Vectors in EN
513
as its representative. That is, u(AC) = v(AB)
+ w(BC).
If h is a real number and D is h of the way from A to B, then hv = v(AD).
We define Ivi as the common length of all the representatives of v. Let (x) be a coordinate system on EN. We define the ith unit point I; by the coordinates (0, 0, ... , 0, 1, 0, ... , 0) where all the coordinates are zero except the ith which is 1. The vector e; is th~ vector of length 1 which has as representative the directed line segment OI;, where 0 is the origin of (x). Theorem A.18. Let (x) be a coordinate system on EN. If A and Bare any points in EN, then N
v(AB) =
L (xf -
xf)e;.
i=l
We prove the result for N = 3, the proof in the general case being --+ similar. The vector v(AB) has a representative OP where P has coordinates PROOF.
(x:- xf, xf- xt, x:- xf).
Let R 1 , R 2 , R 3 , Q1 , Q2 , Q3 have coordinates
R1
= (x:Q1
R 2 = (0, xf- xt, 0),
xf, 0, 0),
= R1 ,
Q2
R 3 = (0, 0,
= (x:- xf, xf- xt, 0),
Q3
x: - xf),
= P.
We observe that R 1 is (x:- xf) of the way from 0 to I 1 ; R 2 is (xf- xt} of the way from 0 to I 2 and R 3 is (x:- xf) of the way from 0 to I 3 • Also,
v(ORt)
= (x:-
xf)e~o
v(OR 2 ) = (xf - xt)e 2 ,
v(OR 3 ) = (x: - xf)e 3 • Hence we see that v(OP) = v(OQd
+ v(Q 1Q2) + v(Q2Q3),
which is the desired result.
D
Theorem A.19. Suppose that e 1 , e 2 , vectors in EN. Suppose that
Then
v + w=
••• ,
eN are mutually perpendicular unit
N
L (a; + b;)e;, i=l lvl
=
N
hv
=
Ja~ +···+a~.
L ha;e;, i=l
Appendixes
514
Let (x) be a coordinate system on EN with origin 0 and unit points I; where v(OI;) = e;. The hypotheses of Theorem A.18 hold. Let A, B, C be the points with coordinates (a 1 , ... , aN), (b 1 , ••. , bN), (ha 1 , ••• , haN). Then Cis h of the way from 0 to A and
PROOF.
v = v(OA),
w = v(AB),
hv
= v(OC),
v
+ w = v(OB).
D
The next theorem is a direct consequence of the above results on vectors. Theorem A.20 (a) The operation of addition for vectors satisfies Axioms A-1 through
A-5 for
addition of real numbers (Chapter 1). (b) Let v and w be two vectors and c, d real numbers. Then (c
+ d)v = cv + dv,
c(v
O·v = 0,
1 · v = v,
c(dv) = (cd)v,
+ w) = cv + cw, (-1)v = -v.
Definition. Let u and v be vectors in EN. The inner (or scalar) product of u and v, denoted u · v, is defined by u·v = Hlu
+ vl 2 -lul 2 -lvl 2 ].
The value of the inner product is independent of the coordinate system. Theorem A.ll. Let u and v be vectors in EN and (x) a coordinate system on EN. Suppose e 1 , ... , eN is a set of mutually orthogonal unit vectors and N
u
=
L a;e;, i=l
Then N
u·v = ~ L..t a.b.· 1 ,, i=l
also, u·v
= v·u,
where '!:Y is any vector in EN.
u · (cv
+ dw) = c(u · v) + d(u · w),
Answers to Odd-Numbered Problems
Section 1.2 1. The inverse ofT is a function if the a; are all distinct.
7. (a) Proposition 1.3. (b) Propositions 1.4, 1.6, 1.7. (c)
f "f i!j! (n-n! i - j)!
a•-i-ibici
i=O j=O
11. Yes
2
3
X
0
1
2
3
0
0
1
0
3
2
2
2
3
0
1
3
3
2
1
0
+
0
0
0
2
3
0
0
0
0
1
2
3
2
0
2
3
1
3
0
3
1
2
Section 1.3 1. Yes. Each number a
5. (0, oo) 13. ( -1, 5)
+ bj7, a, b rational corresponds to a unique point on the line.
7.(-4,00) 15. (-oo,O)u(-!f,oo)
9.(-2,3)
11.[-1,2) 17. (-oo,l)u(2,oo)
19. (!, 3) Section 1.4
9. (b) Yes. (c) Yes. 20. lfSisasubsetofN x N containing(!, l)such thatforall(m, n)inSboth(m and (m, n + 1) are inS, then S = N x N.
+ 1, n) 515
Answers to Odd-Numbered Problems
516
Section 2.1 1. {) = 0.005
3. {) = 0.02
7. {)
9. {) = 0.005
0.06
=
13. {) = 0.07
5. {) = 0.002 11. {) = 0.01
15. {) = 0.09
17. {) = 0.7
Section 2.2 5. No.
Section 2.3 1. Yes.
3. No. Neither.
5. No. Neither.
7. No. On the right.
9. No. Neither.
11. No. Neither.
13. Limit is 1; limit is 0; limit doesn't exist.
15. If limx-•+ / 1 (x) = L 1 , limx-•+ / 2 (x) = L 2 limx-•+ g(x) = L 1 + L 2 •
and
g(x)
= / 1 (x) + / 2 (x),
then
17. Suppose limx-•- f(x) = L, Iimx-•- g(x) = M and f(x) ::;; g(x) for all x in an interval with a as right endpoint. Then L ::;; M.
Section 2.4 1. 0
3. +oo
7. +oo
9. 2
5. 0
13. Suppose that/ and g are functions on IR 1 to IR 1 • Iff is continuous at Land g(x)-+ L as x-+ -oo, then limx--oof[g(x)] = f(L).
g(x) = -x; 17. Examples are (a) f(x) = 2x, (b) f(x) = x, g(x) = -2x; (c) f(x) = x +A, g(x) = -x. 19. Suppose that limx_ +oo f(x) = L, limx_ +oo g(x) = M. If f(x) ::;; g(x) for all x > A for some constant A, then L ::;; M.
Section 2.5 1. (a)
Xn
= (n + t)n; (b) Yn =
3. Hint: Write a" = (1
+ (a -
11. 1
2nn;
(c)
Zn
= (2n + 1)1t.
1))" and use the binomial theorem. 13. 0
Section 3.1 3. If a0 < 0, then f(x)-+ +oo as x-+ ±oo; if a0 > 0, then f(x)-+ -oo as x-+ ±oo. 7. f(x) = x sin(1/x), c = 0. Section 3.2
1. 3, yes; 0, no.
3. 3, no;
-1, no.
Answers to Odd-Numbered Problems
5. 1, no;
t, yes.
517
7. 31t/2, yes;
1t/2, yes.
11. (b) inf S = inf{bi} provided the right-hand expression exists. Similarly, supS= supi{Bi}·
19. Example 1: Take Ii = [(i - 1)1t, i1t], f(x) = cos x; Example 2: 11 = [ -1, 0], Ii = [1/i1t, 1/(i - 1)1t], f(x) = 0, x x>O.
~
O,f(x) = cos(1/x),
Section 3.3
1. No; {x2.}.
3. No; {x2.}.
5. Yes.
7. No; {x 3 .}.
9. (a) xk,n = {k + 1/n}, k = 1, 2, ... , N; n = 1, 2, ... . (b) xk,n={k+1/n}, k=1,2,3, ... ,n=1,2,3, ... . Section 3.5 3. f(x)
= xf(x + 1)
Section 3.7
1. Yes.
3. Yes.
5. No; {x 3.}.
13. (b) A finite subfamily has a smallest interval IN. The point x = 1/(N + 3) is not covered.
15. No. 17. Finite subfamilies which cover E are 1413 and 1312 • Section 4.1
5. Let J, u, v be functions on IR 1 such that v has a derivative at x 0 , u has a derivative at v(x 0 ), and f has a derivative at u[v(x 0 )]. 9. (b) rJ(h) = 3 -
3 + 3h (1
+ h2 + h)3
15. For k ~ (n - 1)/2
21.
1
23. e
Section 4.2
1. 11 = ( -oo, -1], 12 = [ -1, +oo), J1 = [1, +oo), g 1 (x) = Jx"=l-1
-Jx"=i- 1, g (x) = 2
3. 11 = (-oo, 2], 12 = [2, +oo), J1 = ( -oo, 4], J2 = ( -oo, 4], g 1 (x) = 2- ~.
g 2 (x)=2+~ 5. 11 = ( -oo, -2), 12 = ( -2, +oo), J 1 = (2, +oo), J2 = ( -oo, 2), g 1 (x) = 2x/(2- x), g 2 (x) = 2x/(2- x)
7. 11 = (-oo, -1], 12 = [ -1, +1], 13 = [1, +oo), J1 = [ -2, 0), J2 = [ -2, 2],J3 = (0, 2],g 1 (x) = (2 + x 2 )/x,g 2 (x) = (2x 2)/x,g 3 (x) = (2 + x 2)/x
J4-
J4-
J4-
Answers to Odd-Numbered Problems
518 9. / 1 = (-oo, +oo),J1 = (-oo, +oo)
11. / 1 = (-oo, -3), 12 = [ -3, 1], / 3 = [1, +oo) J 1 = ( -oo, 31], J2 = [ -1, 31], J3 = [31, +oo) 15. f'(x) = 3(x - 1) 2 , g'(x) =
lx- 213
17. f'(x) =cos x, g'(x) = 1/JI="? Section 5.1
1. s+u, ~)
= -H; s_(f, ~) =
9. Note that
Li'=
1
z6s
[f(x;)- f(x;-d] - f(x.)- f(x 0 ) = f(b)- f(a). 19. Hint: Use the result of Problem 18.
15. Yes. Section 5.2
1. (c) Choose f(x) = x, a :::; x < b, f(b) = b - 1. 3. Note that d(uv) = u dv
+ v du.
5. For 0 :::; x :::; 1let f(x) = 1 for x rational, f(x) = -1 for x irrational. Section 6.1
7. No.
5. Yes.
3. Not equivalent.
11. Either all X; are zero or there is a number 2 such that Y; = h; fori = 1, 2, ... , n, ... . Section 6.2
3. The square with vertices at (1, 0), (0, 1), ( -1, 0), (0, -1). 5. No. The set {1/n }, n = 1, 2, ... , is an infinite set of isolated points. 7. A={x:O:::;x:::;1}. 11. The sets A.= {(x, y): 0:::; x 2
+ y2
< 1/n}, n = 1, 2, ....
15.Define (in IR 1 ), A;={x:1/i<x:::;1}. Then {x: 0:::; x:::; 1} =1- U A;= {x: 0 < x:::; 1}.
UA;={x:O<x:::;1};
B=
Section 6.3
1. Arrange the rational points as shown in Figure 6.5. Section 6.4
5. Choose p = 7. Choose x" =
-fo. (x~,
Xz, ... , x;:, ... ) so that xi: =
1 if k = n, xi: = 0 otherwise.
Section 6.5
7. Statement: Let f, g, h be functions defined on a set A in a metric space, and suppose PoE A. Ifj(p):::; g(p):::; h(p)forallp E A andiflimP~PJ(p) = limP~P 0 h(p) = Lfor pEA, then limP~Pog(p) = L for pEA.
Answers to Odd-Numbered Problems
519
9. (a) For every B > 0 there is a 8 > 0 such that d2 (f(p), f(p 0 )) < e whenever d 1 (p, p 0 ) < 8 where d 1 , d2 are the metrics in S1 , S2, respectively. (b) Same as 9(a) except that the points p must belong to A. (c) Foreverye > Othereisa8 > Osuchthatd 2(f(p), q 0 ) < ewheneverd 1 (p, p 0 ) < 8 and pEA, where d 1 , d2 are the metrics of S1 , S2 respectively. 11. Let A= 11 u 12 , 11 = {x: 0 ~ x ~ 1}, 12 = {x: 2 ~ x ~ 3}. Define f = 1 on 12 •
f =0
on 11 ,
Section 7.1
5. H, 3(x)
=
-4x 3(xf
+ x~) + 2xf(xfx 3 + x 4 ) + 2[cos(x 1 + x 3)- 2x 4 ] sin(x 1 + x 3)
Section 7.2
7. f(x) = f(a)
+ Dtf(a)(x 1 - a 1 ) + Dd(a)(x 2 - a 2) + DJI(a)(x 3 - a 3)
ad 2 + D1 Dd(a)(x 1 - a 1 )(x 2 - a 2) + DtD3f(a)(xt- ad(x3- a3) + tDif(a)(x2- a2)2 + D2DJf(a)(x 2 - a 2)(x 3 - a 3) + !Dif(a)(x 3 - a 3)2
+ !DU(a)(x 1
-
+ iDif(~)(xt - ad 3 + !DfDd(~)(xt - ad 2(x2- a2) + !DfDJf(~)(xt- at) 2(x3- a3) + !D 1 Dif(~)(xt- ad(x 2 - a 2)2 + D 1 D2 D3 f(~)(x 1 - a 1 )(x 2 - a 2)(x 3 - a 3) + !DtDif(~)(xt - at)(x 3 - a3) 2 + iDif(~)(x 2 - a 2)3 + !D1DJf(~)(x2 - a2) 2(x3 - a3) + !D2 D~f(~)(x2 - a 2)(x 3 - a 3)2 + iDU(~)(x3 - a3) 3 11. Positive definite. 13. Negative definite. Section 7.3
11. (a) Definition: The derivative off at a is the lmear function L: lim dM(f(x), L(x)) dN(x, a) .. ~.
=
~N-+
IRM such that
O.
Section 8.1
1. S = {x: 0 ~xi~ 1, xi is rational, i = 1, 2, ... , N}. 3. The result does not hold even in ~ 2 with S
= {(x 1 , x 2): t ~xi~ j, i = 1, 2}.
7. sl = {x: 0 ~xi~ 1, xi rational}, s2 = {x: 0 ~xi~ 1, xi irrational}. sl u s2 is a figure, S1 n S2 = 0. Also S1 - S2 = 0; hence all are figures.
Answers to Odd-Numbered Problems
520 Section 8.2
1. Let F = {x: 0 ~X;~ 1, i = 1, 2, ... , N}. Definef: F-+ IR 1 such thatf(x) is rational and f(x) = 1 if x 1 is irrational.
= Oifx 1
Section 8.3
3. Let F 1 be a regular figure in IRN and G a figure with zero N-dimensional volume. Define F = F 1 v G and suppose f: F-+ IR 1 is Riemann integrable on F1 and unbounded on G. 7. (b) Observe that for every subdivision into subsquares of S, we have s+(f, ~) = 1 and s-(f, ~) = 0. Section 9.1
1. Convergent.
3. Convergent.
7. Convergent.
9. Convergent.
5. Convergent.
13. p > 1.
17. Divergent. Section 9.2
1. Absolutely convergent.
3. Conditionally convergent.
5. Divergent.
7. Divergent.
9. Conditionally convergent.
11. Conditionally convergent.
13. Divergent.
15. Convergent.
17. -1 <X< 1
19.
21. 1 <X< 5
23.
-i ~X< i -i <X< -t
co co x" 29. (i) L n! x"; (ii) L 1 . n=O n=O n.
Section 9.3
1. Uniform.
3. Uniform.
7. Not uniform.
9. Uniform.
5. Uniform.
Section 9.4
1. h < 1
3. h < 1
7. h < 1
9. h < 1
23. 1 +
f n=l
251
(-1/2)(-3/2) ... (-1/2- n + 1)x2n, lxl < 1 n!
~(-1r(-3)(-4) ... (-3-n+1)
• + n=l L...
5. h < 3/2
I
n.
n X '
jxj < 1
Answers to Odd-Numbered Problems
27. 1 +
f n=l
521
(-1t(-3)(-4) ... (-3- n + 1)x2•, lxl < 1 n!
~ (-1f(-1/2)(-3/2) ... (-1/2-n+ 1) 6 •+ 3 3 29.x+L.. x,
n!(2n + 1)
n=l
31. 0.47943
33. 0.89837
35. 1.99476
37. 0.23981
lxl < 1
Section 9.5 5. No.
3. No.
Section 9.6 3. No.
9. No. In fact, the sum does not converge. 13. No. In fact, the sum does not converge.
Section 9.7 x2 y2 x3 xy2 1 1+x+---+--. 2 2 6 2 x2 y2 x3 xy2 3. 1 - x + - + - - - - 2 2 6 2
5. 1
Section 10.1 1.
1
2+
2 [sin x
sin 3x
~ -1- + -3- + ... +
11: 2 4[ 3• 3+
cos x
cos 2x
J
sin(2k + 1)x 2k + 1 + . . . '
cos 3x
-~+~-~+···+
-1t~X~1t.
J
( -1)k cos kx k2 +··· '
5 2 4 [cos 2x cos 4x . . . ( -1)k+l cos 2kx] • ~ + ~ -3- - l S + + 4k2 - 1 '
-1t~X~1t.
-11: ~X~ 11:.
2 2 2 2 7 e "- e- " e "- e- "[-2 cos x sin x 2 cos 2x 2 sin 2x ( -1)"2 cos kx • 411: + 11: 5 + -5- + - - 8 - - --8- + k2 + 4
_(-1}"ksinkx+···] k2 + 4 '
-n~x~n.
1 cos 2x 9• 2--24 00 1 11. - L -2 - - sin(2n- 1)x, 11: n=l n- 1
-11: ~ x ~ 11:.
522 13.
Answers to Odd-Numbered Problems
1cos x + ;i cos 3x,
3 19 (a) n . 3
- n ~ x ~ n.
4 cos x 2 sin x 4 cos 2x _ 2 sin 2x . . . 4( -1)k cos kx 12 + 1 + 22 2 + + k2
_
2( -1)k sin kx k
-
+ ···, -
n
~
~
x
n.
Section 10.2
J
1. ~ + ~[cos x _ cos 3x + cos 5x + ... 2 n 1 3 5 '
~
x
~
n.
3 ~_~[cos x cos 3x cos 5x ···] 0 ~ x · 2 n 12 + 32 + 52 + '
~
n.
0
J
8 [sin 2x sin 6x sin lOx 5. -+ - - + - - + - - + · · · 0~x n 2 2 2 '
J
sin x sin 2x sin 3x 7. 2 [ - 1- - -2- + -3- - · · · ,
~
n.
O~x~n.
J
4[ . 1 sin !nx sin fnx 9. -sm-nx+---+---+··· n 2 3 5 ' 11.
4 [cos xn cos 2nx cos 3nx 31 + n2 -1-2- + _2_2_ - _3_2_ +
Section 10.3
J J
sin x sin 2x sin 3x ... 1. 4 [ 1"3-2J+~, 3.
x sin 3x sin 5x ... -n4 [sin 1"3+~+~+
5. f(x) ~
2
...
,
~
J
4 [cos 2x cos 4x cos 6x 22- 1 - 42- 1 + 62- 1 +... '
-n; ~X~ n.
sin 6x J -n4 [ 2(2sin 2x1) + 4(4sin 4x1) + 6(62 - 1) + · · · ' 2 -
n+t
11.~ n Section 11.1
1 1
0
cos xt --dt 1+t
3. 1/(x + 1), x =1- -1
n.
-n~x~n.
2 -
a. 0, I